Same Fringe

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Task
Same Fringe
You are encouraged to solve this task according to the task description, using any language you may know.

Write a routine that will compare the leaves ("fringe") of two binary trees to determine whether they are the same list of leaves when visited left-to-right. The structure or balance of the trees does not matter; only the number, order, and value of the leaves is important.

Any solution is allowed here, but many computer scientists will consider it inelegant to collect either fringe in its entirety before starting to collect the other one. In fact, this problem is usually proposed in various forums as a way to show off various forms of concurrency (tree-rotation algorithms have also been used to get around the need to collect one tree first). Thinking of it a slightly different way, an elegant solution is one that can perform the minimum amount of work to falsify the equivalence of the fringes when they differ somewhere in the middle, short-circuiting the unnecessary additional traversals and comparisons.

Any representation of a binary tree is allowed, as long as the nodes are orderable, and only downward links are used (for example, you may not use parent or sibling pointers to avoid recursion).

Contents

[edit] Ada

We first specify a "Bin_Trees" package with standard subprograms to handle binary trees. The package is generic, which allows Data to be essentially of any type.

generic
type Data is private;
package Bin_Trees is
 
type Tree_Type is private;
 
function Empty(Tree: Tree_Type) return Boolean;
function Left (Tree: Tree_Type) return Tree_Type;
function Right(Tree: Tree_Type) return Tree_Type;
function Item (Tree: Tree_Type) return Data;
function Empty return Tree_Type;
 
procedure Destroy_Tree(N: in out Tree_Type);
function Tree(Value: Data) return Tree_Type;
function Tree(Value: Data; Left, Right : Tree_Type) return Tree_Type;
 
private
 
type Node;
type Tree_Type is access Node;
type Node is record
Left, Right: Tree_Type := null;
Item: Data;
end record;
 
end Bin_Trees;

The implementation is straightforward.

with Ada.Unchecked_Deallocation;
 
package body Bin_Trees is
 
function Empty(Tree: Tree_Type) return Boolean is
begin
return Tree = null;
end Empty;
 
function Empty return Tree_Type is
begin
return null;
end Empty;
 
function Left (Tree: Tree_Type) return Tree_Type is
begin
return Tree.Left;
end Left;
 
function Right(Tree: Tree_Type) return Tree_Type is
begin
return Tree.Right;
end Right;
 
function Item (Tree: Tree_Type) return Data is
begin
return Tree.Item;
end Item;
 
procedure Destroy_Tree(N: in out Tree_Type) is
procedure free is new Ada.Unchecked_Deallocation(Node, Tree_Type);
begin
if not Empty(N) then
Destroy_Tree(N.Left);
Destroy_Tree(N.Right);
Free(N);
end if;
end Destroy_Tree;
 
function Tree(Value: Data; Left, Right : Tree_Type) return Tree_Type is
Temp : Tree_Type := new Node;
begin
Temp.all := (Left, Right, Value);
return Temp;
end Tree;
 
function Tree(Value: Data) return Tree_Type is
begin
return Tree(Value, null, null);
end Tree;
 
end Bin_Trees;

Next, we specify and implement package that defines a task type for tree traversal. This allows us to run any number of tree traversals in parallel, even on the same tree.

   generic
with procedure Process_Data(Item: Data);
with function Stop return Boolean;
with procedure Finish;
package Bin_Trees.Traverse is
task Inorder_Task is
entry Run(Tree: Tree_Type);
-- this will call each Item in Tree and, at the very end, it will call Finish
-- except when Stop becomes true; in this case, the task terminates
end Inorder_Task;
end Bin_Trees.Traverse;
   package body Bin_Trees.Traverse is
task body Inorder_Task is
procedure Inorder(Tree: Tree_Type) is
begin
if not Empty(Tree) and not Stop then
Inorder(Tree.Left);
if not Stop then
Process_Data(Item => Tree.Item);
end if;
if (not Stop) then
Inorder(Tree.Right);
end if;
end if;
end Inorder;
T: Tree_Type;
begin
accept Run(Tree: Tree_Type) do
T := Tree;
end Run;
Inorder(T);
Finish;
end Inorder_Task;
end Bin_Trees.Traverse;

When comparing two trees T1 and T2, we will define two tasks, a "Producer.Inorder_Task" and a "Consumer.Inorder_Task". The producer will write data items to a buffer, the consumer will read items from the buffer and compare them with its own data items. Both tasks will terminate when the consumer finds a data item different from the one written by the producer, or when either task has written its last item and the other one has items left.

A third auxiliary task just waits until the consumer has finished and the result of the fringe comparison can be read.

with Ada.Text_IO, Bin_Trees.Traverse;
 
procedure Main is
 
package B_Trees is new Bin_Trees(Character); use B_Trees;
 
function Same_Fringe(T1, T2: Tree_Type) return Boolean is
 
protected type Buffer_Type is
entry Write(Item: Character);
entry Write_Done;
entry Read_And_Compare(Item: Character);
entry Read_Done;
entry Wait_For_The_End;
function Early_Abort return Boolean;
function The_Same return Boolean;
private
Current: Character;
Readable: Boolean := False;
Done: Boolean := False;
Same: Boolean := True;
Finished: Boolean := False;
end Buffer_Type;
 
protected body Buffer_Type is
 
entry Write(Item: Character) when not Readable is
begin
Readable := True;
Current  := Item;
end Write;
 
entry Write_Done when not Readable is
begin
Readable := True;
Done  := True;
end Write_Done;
 
entry Read_And_Compare(Item: Character) when Readable is
begin
if Done then -- Producer is already out of items
Same := False;
Finished := True;
-- Readable remains True, else Consumer might lock itself out
elsif
Item /= Current then
Same := False;
Finished := True;
Readable := False;
else
Readable := False;
end if;
end Read_And_Compare;
 
entry Read_Done when Readable is
begin
Readable := False;
Same  := Same and Done;
Finished := True;
end Read_Done;
 
entry Wait_For_The_End when (Finished) or (not Same) is
begin
null; -- "when ..." is all we need
end Wait_For_The_End;
 
function The_Same return Boolean is
begin
return Same;
end The_Same;
 
function Early_Abort return Boolean is
begin
return not The_Same or Finished;
end Early_Abort;
 
end Buffer_Type;
 
Buffer: Buffer_Type;
 
-- some wrapper subprogram needed to instantiate the generics below
 
procedure Prod_Write(Item: Character) is
begin
Buffer.Write(Item);
end Prod_Write;
 
function Stop return Boolean is
begin
return Buffer.Early_Abort;
end Stop;
 
procedure Prod_Stop is
begin
Buffer.Write_Done;
end Prod_Stop;
 
procedure Cons_Write(Item: Character) is
begin
Buffer.Read_And_Compare(Item);
end Cons_Write;
 
procedure Cons_Stop is
begin
Buffer.Read_Done;
end Cons_Stop;
 
package Producer is new B_Trees.Traverse(Prod_Write, Stop, Prod_Stop);
package Consumer is new B_Trees.Traverse(Cons_Write, Stop, Cons_Stop);
 
begin
Producer.Inorder_Task.Run(T1);
Consumer.Inorder_Task.Run(T2);
Buffer.Wait_For_The_End;
return Buffer.The_Same;
end Same_Fringe;
 
procedure Show_Preorder(Tree: Tree_Type; Prefix: String := "") is
use Ada.Text_IO;
begin
if Prefix /= "" then
Ada.Text_IO.Put(Prefix);
end if;
if not Empty(Tree) then
Put("(" & Item(Tree)); Put(", ");
Show_Preorder(Left(Tree)); Put(", ");
Show_Preorder(Right(Tree)); Put(")");
end if;
if Prefix /= "" then
New_Line;
end if;
end Show_Preorder;
 
T_0: Tree_Type := Tree('a', Empty, Tree('b'));
T: array(1 .. 5) of Tree_Type;
 
begin
T(1) := Tree('d', Tree('c'), T_0);
T(2)  := Tree('c', Empty, Tree('a', Tree('d'), Tree('b')));
T(3)  := Tree('e', T(1), T(2));
T(4)  := Tree('e', T(2), T(1));
T(5)  := Tree('e', T_0, Tree('c', Tree('d'), T(1)));
 
-- First display the trees you have (in preorder)
for I in T'Range loop
Show_Preorder(T(I), "Tree(" & Integer'Image(I) & " ) is ");
end loop;
Ada.Text_IO.New_Line;
 
-- Now compare them, which have the same fringe?
for I in T'Range loop
for J in T'Range loop
if Same_Fringe(T(J), T(I)) then
Ada.Text_IO.Put("same(");
else
Ada.Text_IO.Put("DIFF(");
end if;
Ada.Text_IO.Put(Integer'Image(I) & "," & Integer'Image(J) & " ); ");
end loop;
Ada.Text_IO.New_Line;
end loop;
end Main;

Note that we do not call Destroy_Tree to reclaim the dynamic memory. In our case, this is not needed since the memory will be reclaimed at the end of Main, anyway.

Output:
Tree( 1 ) is (d, (c, , ), (a, , (b, , )))
Tree( 2 ) is (c, , (a, (d, , ), (b, , )))
Tree( 3 ) is (e, (d, (c, , ), (a, , (b, , ))), (c, , (a, (d, , ), (b, , ))))
Tree( 4 ) is (e, (c, , (a, (d, , ), (b, , ))), (d, (c, , ), (a, , (b, , ))))
Tree( 5 ) is (e, (a, , (b, , )), (c, (d, , ), (d, (c, , ), (a, , (b, , )))))

same( 1, 1 ); same( 1, 2 ); DIFF( 1, 3 ); DIFF( 1, 4 ); DIFF( 1, 5 ); 
same( 2, 1 ); same( 2, 2 ); DIFF( 2, 3 ); DIFF( 2, 4 ); DIFF( 2, 5 ); 
DIFF( 3, 1 ); DIFF( 3, 2 ); same( 3, 3 ); same( 3, 4 ); DIFF( 3, 5 ); 
DIFF( 4, 1 ); DIFF( 4, 2 ); same( 4, 3 ); same( 4, 4 ); DIFF( 4, 5 ); 
DIFF( 5, 1 ); DIFF( 5, 2 ); DIFF( 5, 3 ); DIFF( 5, 4 ); same( 5, 5 ); 

[edit] Bracmat

( ( T
=
. ( next
= node stack rhs
.  !arg:%?node ?stack
& whl
' ( !node:(?node.?rhs)
& !rhs !stack:?stack
)
& (!node.!stack)
)
& !arg:(?stackA,?stackB)
& whl
' ( !stackA:~
& !stackB:~
& next$!stackA:(?leafA.?stackA)
& next$!stackB:(?leafB.?stackB)
& !leafA:!leafB
)
& out$!arg
& (  !stackA:!stackB:
& !leafA:!leafB
& out$equal
| out$"not equal"
)
)
& T$(x,x)
& T$((x.y),(x.y))
& T$(((x.y).z),(x.y.z))
& T$((x.y.z),(x.y.q))
& T$((x.y),(x.y.q))
& T$((x.y.z),(x.y))
& T$(((x.y).z),(x.z.y))
& T
$ ( (a.b.c.(x.y).z)
, (((a.b).c).x.y.z)
)
);

Output:

x,x
equal
(x.y),(x.y)
equal
((x.y).z),(x.y.z)
equal
(x.y.z),(x.y.q)
not equal
(x.y),(x.y.q)
not equal
(x.y.z),(x.y)
not equal
((x.y).z),(x.z.y)
not equal
  (a.b.c.(x.y).z)
, (((a.b).c).x.y.z)
equal

[edit] C

With rudimentary coroutine support based on ucontext. I don't know if it will compile on anything other than GCC.

#include <stdio.h>
#include <stdlib.h>
#include <ucontext.h>
 
typedef struct {
ucontext_t caller, callee;
char stack[8192];
void *in, *out;
} co_t;
 
co_t * co_new(void(*f)(), void *data)
{
co_t * c = malloc(sizeof(*c));
getcontext(&c->callee);
c->in = data;
 
c->callee.uc_stack.ss_sp = c->stack;
c->callee.uc_stack.ss_size = sizeof(c->stack);
c->callee.uc_link = &c->caller;
makecontext(&c->callee, f, 1, (int)c);
 
return c;
}
 
void co_del(co_t *c)
{
free(c);
}
 
inline void
co_yield(co_t *c, void *data)
{
c->out = data;
swapcontext(&c->callee, &c->caller);
}
 
inline void *
co_collect(co_t *c)
{
c->out = 0;
swapcontext(&c->caller, &c->callee);
return c->out;
}
 
// end of coroutine stuff
 
typedef struct node node;
struct node {
int v;
node *left, *right;
};
 
node *newnode(int v)
{
node *n = malloc(sizeof(node));
n->left = n->right = 0;
n->v = v;
return n;
}
 
void tree_insert(node **root, node *n)
{
while (*root) root = ((*root)->v > n->v)
? &(*root)->left
: &(*root)->right;
*root = n;
}
 
void tree_trav(int x)
{
co_t *c = (co_t *) x;
 
void trav(node *root) {
if (!root) return;
trav(root->left);
co_yield(c, root);
trav(root->right);
}
 
trav(c->in);
}
 
int tree_eq(node *t1, node *t2)
{
co_t *c1 = co_new(tree_trav, t1);
co_t *c2 = co_new(tree_trav, t2);
 
node *p = 0, *q = 0;
do {
p = co_collect(c1);
q = co_collect(c2);
} while (p && q && (p->v == q->v));
 
co_del(c1);
co_del(c2);
return !p && !q;
}
 
int main()
{
int x[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -1 };
int y[] = { 2, 5, 7, 1, 9, 0, 6, 4, 8, 3, -1 };
int z[] = { 0, 1, 2, 3, 4, 5, 6, 8, 9, -1 };
 
node *t1 = 0, *t2 = 0, *t3 = 0;
 
void mktree(int *buf, node **root) {
int i;
for (i = 0; buf[i] >= 0; i++)
tree_insert(root, newnode(buf[i]));
}
 
mktree(x, &t1); // ordered binary tree, result of traversing
mktree(y, &t2); // should be independent of insertion, so t1 == t2
mktree(z, &t3);
 
printf("t1 == t2: %s\n", tree_eq(t1, t2) ? "yes" : "no");
printf("t1 == t3: %s\n", tree_eq(t1, t3) ? "yes" : "no");
 
return 0;
}

[edit] Clojure

fringe-seq produces a lazy sequence of the fringe values of a tree. It's patterned after the standard function tree-seq: aside from the tree to walk, it takes 3 function arguments to handle a general tree structure. branch? returns true for branch nodes -- nodes which could have children, whether they actually do or not. children returns the children of a branch node; content returns the content of a branch node. A fringe value is either the content of a branch node without children, or a non-branch node.

(defn fringe-seq [branch? children content tree]
(letfn [(walk [node]
(lazy-seq
(if (branch? node)
(if (empty? (children node))
(list (content node))
(mapcat walk (children node)))
(list node))))]
(walk tree)))

For this problem, binary trees are represented as vectors, whose nodes are either [content left right] or just content.

(defn vfringe-seq [v] (fringe-seq vector? #(remove nil? (rest %)) first v))
(println (vfringe-seq [10 1 2])) ; (1 2)
(println (vfringe-seq [10 [1 nil nil] [20 2 nil]])) ; (1 2)

Then we can use a general sequence-equality function:

(defn seq= [s1 s2]
(cond
(and (empty? s1) (empty? s2)) true
(not= (empty? s1) (empty? s2)) false
(= (first s1) (first s2)) (recur (rest s1) (rest s2))
 :else false))

[edit] D

[edit] Short Version

A short recursive solution that is not lazy (and lacks the const/immutable/pure/nothrow):

struct Node(T) {
T data;
Node* L, R;
}
 
bool sameFringe(T)(Node!T* t1, Node!T* t2) {
T[] scan(Node!T* t) {
if (!t) return [];
return (!t.L && !t.R) ? [t.data] : scan(t.L) ~ scan(t.R);
}
return scan(t1) == scan(t2);
}
 
void main() {
import std.stdio;
alias N = Node!int;
auto t1 = new N(10, new N(20, new N(30, new N(40), new N(50))));
auto t2 = new N(1, new N(2, new N(3, new N(40), new N(50))));
writeln(sameFringe(t1, t2));
auto t3 = new N(1, new N(2, new N(3, new N(40), new N(51))));
writeln(sameFringe(t1, t3));
}
Output:
true
false

[edit] Strong Lazy Version

This version is quite long because it tries to be reliable. The code contains contracts, unit tests, annotations, and so on.

import std.array: empty;
import std.algorithm: equal;
 
 
// Replace with an efficient stack when available in Phobos.
struct Stack(T) {
private T[] data;
 
public @property bool empty() const pure nothrow {
return data.empty;
}
 
// Can't be const if T isn't a value or const.
public @property T head() const pure nothrow
in {
assert(!data.empty);
} body {
return data[$ - 1];
}
 
public void push(T x) pure nothrow {
data ~= x;
}
 
public void pop() pure nothrow
in {
assert(!data.empty);
} body {
data.length--;
}
}
 
 
struct BinaryTreeNode(T) {
T data;
BinaryTreeNode* left, right;
}
 
 
struct Fringe(T) {
alias const(BinaryTreeNode!T)* BT;
private Stack!BT stack;
 
pure nothrow invariant {
assert(stack.empty || isLeaf(stack.head));
}
 
public this(BT t) pure nothrow {
if (t != null) {
stack.push(t);
if (!isLeaf(t)) {
// Here the invariant doesn't hold.
// invariant isn't called for private methods.
nextLeaf;
}
}
}
 
public @property bool empty() const pure nothrow {
return stack.empty;
}
 
public @property T front() const pure nothrow
in {
assert(!stack.empty && stack.head != null);
} body {
return stack.head.data;
}
 
public void popFront() pure nothrow
in {
assert(!stack.empty);
} body {
stack.pop();
if (!empty())
nextLeaf();
}
 
private static bool isLeaf(in BT t) pure nothrow {
return t != null && t.left == null && t.right == null;
}
 
private void nextLeaf() pure nothrow
in {
assert(!stack.empty);
} body {
auto t = stack.head;
 
while (!stack.empty && !isLeaf(t)) {
stack.pop();
if (t.right != null)
stack.push(t.right);
if (t.left != null)
stack.push(t.left);
t = stack.head;
}
}
}
 
 
bool sameFringe(T)(in BinaryTreeNode!T* t1, in BinaryTreeNode!T* t2)
pure nothrow {
return Fringe!T(t1).equal(Fringe!T(t2));
}
 
 
unittest {
alias BinaryTreeNode!int N;
 
static N* n(in int x, N* l=null, N* r=null) pure nothrow {
return new N(x, l, r);
}
 
{
N* t;
assert(sameFringe(t, t));
}
 
{
const t1 = n(10);
const t2 = n(10);
assert(sameFringe(t1, t2));
}
 
{
const t1 = n(10);
const t2 = n(20);
assert(!sameFringe(t1, t2));
}
 
{
const t1 = n(10, n(20));
const t2 = n(30, n(20));
assert(sameFringe(t1, t2));
}
 
{
const t1 = n(10, n(20));
const t2 = n(10, n(30));
assert(!sameFringe(t1, t2));
}
 
{
const t1 = n(10, n(20), n(30));
const t2 = n(5, n(20), n(30));
assert(sameFringe(t1, t2));
}
 
{
const t1 = n(10, n(20), n(30));
const t2 = n(5, n(20), n(35));
assert(!sameFringe(t1, t2));
}
 
{
const t1 = n(10, n(20, n(30)));
const t2 = n(1, n(2, n(30)));
assert(sameFringe(t1, t2));
}
 
{
const t1 = n(10, n(20, n(30, n(40), n(50))));
const t2 = n(1, n(2, n(3, n(40), n(50))));
assert(sameFringe(t1, t2));
}
 
{
const t1 = n(10, n(20, n(30, n(40), n(50))));
const t2 = n(1, n(2, n(3, n(40), n(51))));
assert(!sameFringe(t1, t2));
}
}
 
 
void main() {
import std.stdio;
alias N = BinaryTreeNode!int;
 
static N* n(in int x, N* l=null, N* r=null) pure nothrow {
return new N(x, l, r);
}
 
const t1 = n(10, n(20, n(30, n(40), n(50))));
writeln("fringe(t1): ", Fringe!int(t1));
 
const t2 = n(1, n(2, n(3, n(40), n(50))));
writeln("fringe(t2): ", Fringe!int(t2));
 
const t3 = n(1, n(2, n(3, n(40), n(51))));
writeln("fringe(t3): ", Fringe!int(t3));
 
writeln("sameFringe(t1, t2): ", sameFringe(t1, t2));
writeln("sameFringe(t1, t3): ", sameFringe(t1, t3));
}
Output:
fringe(t1): [40, 50]
fringe(t2): [40, 50]
fringe(t3): [40, 51]
sameFringe(t1, t2): true
sameFringe(t1, t3): false

[edit] Range Generator Version (Lazy)

import std.stdio, std.concurrency, std.range, std.algorithm;
 
struct Node(T) {
T data;
Node* L, R;
}
 
Generator!T fringe(T)(Node!T* t1) {
return new typeof(return)({
if (t1 != null) {
if (t1.L == null && t1.R == null) // Is a leaf.
yield(t1.data);
else
foreach (data; t1.L.fringe.chain(t1.R.fringe))
yield(data);
}
});
}
 
bool sameFringe(T)(Node!T* t1, Node!T* t2) {
return t1.fringe.equal(t2.fringe);
}
 
void main() {
alias N = Node!int;
 
auto t1 = new N(10, new N(20, new N(30, new N(40), new N(50))));
auto t2 = new N(1, new N(2, new N(3, new N(40), new N(50))));
sameFringe(t1, t2).writeln;
 
auto t3 = new N(1, new N(2, new N(3, new N(40), new N(51))));
sameFringe(t1, t3).writeln;
 
auto t4 = new N(1, new N(2, new N(3, new N(40))));
sameFringe(t1, t4).writeln;
 
N* t5;
sameFringe(t1, t5).writeln;
sameFringe(t5, t5).writeln;
 
auto t6 = new N(2);
auto t7 = new N(1, new N(2));
sameFringe(t6, t7).writeln;
}
Output:
true
false
false
false
true
true

[edit] Go

package main
 
import "fmt"
 
type node struct {
int
left, right *node
}
 
// function returns a channel that yields the leaves of the tree.
// the channel is closed after all leaves are received.
func leaves(t *node) chan int {
ch := make(chan int)
// recursive function to walk tree.
var f func(*node)
f = func(n *node) {
if n == nil {
return
}
// leaves are identified by having no children.
if n.left == nil && n.right == nil {
ch <- n.int
} else {
f(n.left)
f(n.right)
}
}
// goroutine runs concurrently with others.
// it walks the tree then closes the channel.
go func() {
f(t)
close(ch)
}()
return ch
}
 
func sameFringe(t1, t2 *node) bool {
f1 := leaves(t1)
f2 := leaves(t2)
for l1 := range f1 {
// both trees must yield a leaf, and the leaves must be equal.
if l2, ok := <-f2; !ok || l1 != l2 {
return false
}
}
// there must be nothing left in f2 after consuming all of f1.
_, ok := <-f2
return !ok
}
 
func main() {
// the different shapes of the trees is shown with indention.
// the leaves are easy to spot by the int: key.
t1 := &node{3,
&node{1,
&node{int: 1},
&node{int: 2}},
&node{8,
&node{int: 5},
&node{int: 13}}}
// t2 with negative values for internal nodes that can't possibly match
// positive values in t1, just to show that only leaves are being compared.
t2 := &node{-8,
&node{-3,
&node{-1,
&node{int: 1},
&node{int: 2}},
&node{int: 5}},
&node{int: 13}}
fmt.Println(sameFringe(t1, t2)) // prints true.
}

[edit] Haskell

Since Haskell is lazy, simply getting the fringes and comparing them for equality will do. It will only do as much as work as necessary and will stop at the first difference.

To get the fringe, we can simply use the solution for Flatten a list, slightly modified for a binary tree instead of a general tree:

data Tree a = Leaf a | Node (Tree a) (Tree a)
deriving (Show, Eq)
 
fringe :: Tree a -> [a]
fringe (Leaf x) = [x]
fringe (Node n1 n2) = fringe n1 ++ fringe n2
 
sameFringe :: (Eq a) => Tree a -> Tree a -> Bool
sameFringe t1 t2 = fringe t1 == fringe t2
 
main = do
let a = Node (Leaf 1) (Node (Leaf 2) (Node (Leaf 3) (Node (Leaf 4) (Leaf 5))))
b = Node (Leaf 1) (Node (Node (Leaf 2) (Leaf 3)) (Node (Leaf 4) (Leaf 5)))
c = Node (Node (Node (Node (Leaf 1) (Leaf 2)) (Leaf 3)) (Leaf 4)) (Leaf 5)
print $ sameFringe a a
print $ sameFringe a b
print $ sameFringe a c
 
let x = Node (Leaf 1) (Node (Leaf 2) (Node (Leaf 3) (Node (Leaf 4) (Node (Leaf 5) (Leaf 6)))))
y = Node (Leaf 0) (Node (Node (Leaf 2) (Leaf 3)) (Node (Leaf 4) (Leaf 5)))
z = Node (Leaf 1) (Node (Leaf 2) (Node (Node (Leaf 4) (Leaf 3)) (Leaf 5)))
print $ sameFringe a x
print $ sameFringe a y
print $ sameFringe a z
Output:
True
True
True
False
False
False

[edit] Icon and Unicon

The following solution works in both languages:

procedure main()
aTree := [1, [2, [4, [7]], [5]], [3, [6, [8], [9]]]]
bTree := [1, [2, [4, [7]], [5]], [3, [6, [8], [9]]]]
write("aTree and bTree ",(sameFringe(aTree,bTree),"have")|"don't have",
" the same leaves.")
cTree := [1, [2, [4, [7]], [5]], [3, [6, [8]]]]
dTree := [1, [2, [4, [7]], [5]], [3, [6, [8], [9]]]]
write("cTree and dTree ",(sameFringe(cTree,dTree),"have")|"don't have",
" the same leaves.")
end
 
procedure sameFringe(a,b)
return same{genLeaves(a),genLeaves(b)}
end
 
procedure same(L)
while n1 := @L[1] do {
n2 := @L[2] | fail
if n1 ~== n2 then fail
}
return not @L[2]
end
 
procedure genLeaves(t)
suspend (*(node := preorder(t)) == 1, node[1])
end
 
procedure preorder(L)
if \L then suspend L | preorder(L[2|3])
end

Output:

->sf
aTree and bTree have the same leaves.
cTree and dTree don't have the same leaves.
->

[edit] J

sameFringe=: -:&([: ; <S:0)

Note that the time/space optimizations here can change with the language implementation, but current implementations make no effort to treat trees efficiently.

That said, note also that binary trees tend to be a poor data structure choice in J. First, they shift the focus form "what needs to be done" to (in minute detail) "how to do it". This typically means that (for example) combining operations into batches becomes difficult. And, typically, we can find other strategies (some of which have analogies to trees) to achieve the desired efficiencies.

Anyways, here's a recursive routine to convert a flat list into a binary tree:

list2tree=: (<.@-:@# ({. ,&<&list2tree}. ) ])^:(1<#)

And, here are two differently structured trees which represent the same underlying data:

bp=: list2tree p: i.11
ubp=: p:L:0] 10;~list2tree i.10

And, here's our original operation in action (1 {:: ubp is a subtree of ubp which omits a leaf node):

   ubp sameFringe bp
1
bp sameFringe 1 {:: ubp
0

[edit] Java

The code defines a Node interface, an implementation (SimpleNode), and a pair of methods to do the comparison (areLeavesSame and advanceToLeaf). The method simpleWalk() is to show what leaves are present in each tree.

Code:

import java.util.*;
 
class SameFringe
{
public interface Node<T extends Comparable<? super T>>
{
Node<T> getLeft();
Node<T> getRight();
boolean isLeaf();
T getData();
}
 
public static class SimpleNode<T extends Comparable<? super T>> implements Node<T>
{
private final T data;
public SimpleNode<T> left;
public SimpleNode<T> right;
 
public SimpleNode(T data)
{ this(data, null, null); }
 
public SimpleNode(T data, SimpleNode<T> left, SimpleNode<T> right)
{
this.data = data;
this.left = left;
this.right = right;
}
 
public Node<T> getLeft()
{ return left; }
 
public Node<T> getRight()
{ return right; }
 
public boolean isLeaf()
{ return ((left == null) && (right == null)); }
 
public T getData()
{ return data; }
 
public SimpleNode<T> addToTree(T data)
{
int cmp = data.compareTo(this.data);
if (cmp == 0)
throw new IllegalArgumentException("Same data!");
if (cmp < 0)
{
if (left == null)
return (left = new SimpleNode<T>(data));
return left.addToTree(data);
}
if (right == null)
return (right = new SimpleNode<T>(data));
return right.addToTree(data);
}
}
 
public static <T extends Comparable<? super T>> boolean areLeavesSame(Node<T> node1, Node<T> node2)
{
Stack<Node<T>> stack1 = new Stack<Node<T>>();
Stack<Node<T>> stack2 = new Stack<Node<T>>();
stack1.push(node1);
stack2.push(node2);
// NOT using short-circuit operator
while (((node1 = advanceToLeaf(stack1)) != null) & ((node2 = advanceToLeaf(stack2)) != null))
if (!node1.getData().equals(node2.getData()))
return false;
// Return true if finished at same time
return (node1 == null) && (node2 == null);
}
 
private static <T extends Comparable<? super T>> Node<T> advanceToLeaf(Stack<Node<T>> stack)
{
while (!stack.isEmpty())
{
Node<T> node = stack.pop();
if (node.isLeaf())
return node;
Node<T> rightNode = node.getRight();
if (rightNode != null)
stack.push(rightNode);
Node<T> leftNode = node.getLeft();
if (leftNode != null)
stack.push(leftNode);
}
return null;
}
 
public static void main(String[] args)
{
SimpleNode<Integer> headNode1 = new SimpleNode<Integer>(35, new SimpleNode<Integer>(25, new SimpleNode<Integer>(15, new SimpleNode<Integer>(10), new SimpleNode<Integer>(20)), new SimpleNode<Integer>(30)), new SimpleNode<Integer>(45, new SimpleNode<Integer>(40), new SimpleNode<Integer>(50)));
SimpleNode<Integer> headNode2 = new SimpleNode<Integer>(24, new SimpleNode<Integer>(14, new SimpleNode<Integer>(10), new SimpleNode<Integer>(16, null, new SimpleNode<Integer>(20))), new SimpleNode<Integer>(34, new SimpleNode<Integer>(30), new SimpleNode<Integer>(42, new SimpleNode<Integer>(40), new SimpleNode<Integer>(56, new SimpleNode<Integer>(50), null))));
SimpleNode<Integer> headNode3 = new SimpleNode<Integer>(24, new SimpleNode<Integer>(14, new SimpleNode<Integer>(10), new SimpleNode<Integer>(16, null, new SimpleNode<Integer>(20))), new SimpleNode<Integer>(34, new SimpleNode<Integer>(30), new SimpleNode<Integer>(42, new SimpleNode<Integer>(40), new SimpleNode<Integer>(50, null, new SimpleNode<Integer>(56)))));
System.out.print("Leaves for set 1: ");
simpleWalk(headNode1);
System.out.println();
System.out.print("Leaves for set 2: ");
simpleWalk(headNode2);
System.out.println();
System.out.print("Leaves for set 3: ");
simpleWalk(headNode3);
System.out.println();
System.out.println("areLeavesSame(1, 2)? " + areLeavesSame(headNode1, headNode2));
System.out.println("areLeavesSame(2, 3)? " + areLeavesSame(headNode2, headNode3));
}
 
public static void simpleWalk(Node<Integer> node)
{
if (node.isLeaf())
System.out.print(node.getData() + " ");
else
{
Node<Integer> left = node.getLeft();
if (left != null)
simpleWalk(left);
Node<Integer> right = node.getRight();
if (right != null)
simpleWalk(right);
}
}
}

Output:

Leaves for set 1: 10 20 30 40 50
Leaves for set 2: 10 20 30 40 50
Leaves for set 3: 10 20 30 40 56
areLeavesSame(1, 2)? true
areLeavesSame(2, 3)? false

[edit] jq

Works with: jq version 1.4

A binary tree can be conveniently represented in jq as a nested array, e.g. [1,[2,3]]. This is the data structure used by same_fringe(t;u) as defined in this section.

With this data structure, a test for whether two trees, s and t, have the same fringes could be implemented simply as:

(t|flatten) == (s|flatten)

but this entails generating the lists of leaves.

To accomplish the "same fringe" task efficiently in jq 1.4 without generating a list of leaves, a special-purpose function is needed. This special-purpose function, which is here named "next", would ordinarily be defined as an inner function of "same_fringe", but for clarity, it is defined as a top-level function.

# "next" allows one to generate successive leaves, one at a time. This is accomplished
# by ensuring that the non-null output of a call to "next" can also serve as input.
#
# "next" returns null if there are no more leaves, otherwise it returns [leaf, nodes]
# where "leaf" is the next leaf, and nodes is an array of nodes still to be traversed.
# Input has the same form, but on input, "leaf" is ignored unless it is an array.
def next:
def _next:
.[0] as $node | .[1] as $nodes
| if ($node|type) == "array" then
if $node|length != 2 then
error("improper node: \($node) should have 2 items") else . end
| [ $node[0], [$node[1]] + $nodes]
elif $nodes|length > 0 then [$nodes[0], $nodes[1:]]
else null
end;
_next as $n
| if $n == null then null
elif ($n[0]|type) == "array" then $n|next
else $n
end;
 
# t and u must be binary trees
def same_fringe(t;u):
# x and y must be suitable for input to "next"
def eq(x;y):
if x == null then y == null
elif y == null then false
elif x[0] != y[0] then false
else eq( x|next; y|next)
end;
 
eq([t,[]]|next; [u,[]]|next) ;

Example:

  [1,[2,[3,[4,[5,[6,7]]]]]] as $a
| [[[[[[1,2],3],4],5],6],7] as $b
| [[[1,2],3],[4,[5,[6,7]]]] as $c
| [[[1,2],2],4] as $d
| same_fringe($a;$a), same_fringe($b;$b), same_fringe($c;$c),
same_fringe($a;$b), same_fringe($a;$c), same_fringe($b;$c),
same_fringe($a;$d), same_fringe($d;$c), same_fringe($b;$d),
 
same_fringe( ["a",["b",["c",[["x","y"],"z"]]]];
[[["a","b"],"c"],["x",["y","z"]]] )
Output:
$ jq -n -f Same_Fringe.jq
true
true
true
true
true
true
false
false
false
true
 

[edit] OCaml

While we could use a lazy datatype such as Stream for this problem, this example implements the short-circuit behavior (returning on first mismatch) by tracking the parse state.

type 'a btree = Leaf of 'a | BTree of ('a btree * 'a btree)
 
let rec next = function
| [] -> None
| h :: t -> match h with
| Leaf x -> Some (x,t)
| BTree(a,b) -> next (a::b::t)
 
let samefringe t1 t2 =
let rec aux s1 s2 = match (next s1, next s2) with
| None, None -> true
| None, _ | _, None -> false
| Some(a,b), Some(c,d) -> (a=c) && aux b d in
aux [t1] [t2]
 
(* Test: *)
let () =
let u = BTree(Leaf 1, BTree(Leaf 2, Leaf 3)) in
let v = BTree(BTree(Leaf 1, Leaf 2), Leaf 3) in
let w = BTree(BTree(Leaf 3, Leaf 2), Leaf 1) in
let check a b =
print_endline (if samefringe a b then "same" else "different") in
check u v; check v u; check v w;

Output:

same
same
different

[edit] Perl

We use a pair of tree iterators to walk through the trees. So we pick the next leaf from each tree while the leaves are identical. If we've picked the last leaf of both trees simulaneously, then both trees had the "same" fringe. If we find a difference or one of the trees runs out of leaves before the other, we immediately return with a "different" fringe.

The tree iterator is pretty simple: we use array references with index 0 as the left subtree and index 1 holding the right subtree. So as we go down the tree towards the first leaf, we push each right subtree that we will consider later onto the rtree stack. Eventually, we'll hit a leaf and return it. The next time we go into the iterator, we simply pull off the last deferred subtree and continue the process.

 
#!/usr/bin/perl
use strict;
 
my @trees = (
# 0..2 are same
[ 'd', [ 'c', [ 'a', 'b', ], ], ],
[ [ 'd', 'c' ], [ 'a', 'b' ] ],
[ [ [ 'd', 'c', ], 'a', ], 'b', ],
# and this one's different!
[ [ [ [ [ [ 'a' ], 'b' ], 'c', ], 'd', ], 'e', ], 'f' ],
);
 
for my $tree_idx (1 .. $#trees) {
print "tree[",$tree_idx-1,"] vs tree[$tree_idx]: ",
cmp_fringe($trees[$tree_idx-1], $trees[$tree_idx]), "\n";
}
 
sub cmp_fringe {
my $ti1 = get_tree_iterator(shift);
my $ti2 = get_tree_iterator(shift);
while (1) {
my ($L, $R) = ($ti1->(), $ti2->());
next if defined($L) and defined($R) and $L eq $R;
return "Same" if !defined($L) and !defined($R);
return "Different";
}
}
 
sub get_tree_iterator {
my @rtrees = (shift);
my $tree;
return sub {
$tree = pop @rtrees;
($tree, $rtrees[@rtrees]) = @$tree while ref $tree;
return $tree;
}
}
 
Output:
tree[0] vs tree[1]: Same
tree[1] vs tree[2]: Same
tree[2] vs tree[3]: Different

[edit] Perl 6

Works with: Rakudo version #67 ("Bicycle")

Unlike in Perl 5, where => is just a synonym for comma, in Perl 6 it creates a true Pair object. So here we use Pair objects for our "cons" cells, just as if we were doing this in Lisp. We use the gather/take construct to harvest the leaves lazily as the elements are visited with a standard recursive algorithm, using multiple dispatch to differentiate nodes from leaves. The eqv value equivalence is applied to the two lists in parallel.

sub fringe ($tree) {
multi sub fringey (Pair $node) { fringey $_ for $node.kv; }
multi sub fringey ( Any $leaf) { take $leaf; }
 
gather fringey $tree;
}
 
sub samefringe ($a, $b) { fringe($a) eqv fringe($b) }

Testing:

my $a = 1 => 2 => 3 => 4 => 5 => 6 => 7 => 8;
my $b = 1 => (( 2 => 3 ) => (4 => (5 => ((6 => 7) => 8))));
my $c = (((1 => 2) => 3) => 4) => 5 => 6 => 7 => 8;
 
my $x = 1 => 2 => 3 => 4 => 5 => 6 => 7 => 8 => 9;
my $y = 0 => 2 => 3 => 4 => 5 => 6 => 7 => 8;
my $z = 1 => 2 => (4 => 3) => 5 => 6 => 7 => 8;
 
say so samefringe $a, $a;
say so samefringe $a, $b;
say so samefringe $a, $c;
 
say not samefringe $a, $x;
say not samefringe $a, $y;
say not samefringe $a, $z;
Output:
True
True
True
True
True
True

[edit] PicoLisp

This uses coroutines to traverse the trees, so it works only in the 64-bit version.

(de nextLeaf (Rt Tree)
(co Rt
(recur (Tree)
(when Tree
(recurse (cadr Tree))
(yield (car Tree))
(recurse (cddr Tree)) ) ) ) )
 
(de cmpTrees (Tree1 Tree2)
(prog1
(use (Node1 Node2)
(loop
(setq
Node1 (nextLeaf "rt1" Tree1)
Node2 (nextLeaf "rt2" Tree2) )
(T (nor Node1 Node2) T)
(NIL (= Node1 Node2)) ) )
(co "rt1")
(co "rt2") ) )

Test:

: (balance '*Tree1 (range 1 7))
-> NIL
: (for N (5 4 6 3 7 1 2) (idx '*Tree2 N T))
-> NIL
 
: (view *Tree1 T)
7
6
5
4
3
2
1
-> NIL
 
: (view *Tree2 T)
7
6
5
4
3
2
1
-> NIL
 
: (cmpTrees *Tree1 *Tree2)
-> T

[edit] Python

This solution visits lazily the two trees in lock step like in the Perl 6 example, and stops at the first miss-match.

try:
from itertools import zip_longest as izip_longest # Python 3.x
except:
from itertools import izip_longest # Python 2.6+
 
def fringe(tree):
"""Yield tree members L-to-R depth first,
as if stored in a binary tree"""

for node1 in tree:
if isinstance(node1, tuple):
for node2 in fringe(node1):
yield node2
else:
yield node1
 
def same_fringe(tree1, tree2):
return all(node1 == node2 for node1, node2 in
izip_longest(fringe(tree1), fringe(tree2)))
 
if __name__ == '__main__':
a = 1, 2, 3, 4, 5, 6, 7, 8
b = 1, (( 2, 3 ), (4, (5, ((6, 7), 8))))
c = (((1, 2), 3), 4), 5, 6, 7, 8
 
x = 1, 2, 3, 4, 5, 6, 7, 8, 9
y = 0, 2, 3, 4, 5, 6, 7, 8
z = 1, 2, (4, 3), 5, 6, 7, 8
 
assert same_fringe(a, a)
assert same_fringe(a, b)
assert same_fringe(a, c)
 
assert not same_fringe(a, x)
assert not same_fringe(a, y)
assert not same_fringe(a, z)
Output:

There is no output, which signifies success.

[edit] Racket

The following is a slight modification of the same-fringe? program from Dorai Sitaram's 1993 PLDI paper titled "Handling Control". This solution uses the fcontrol delimited continuation operator.

 
#lang racket
(require racket/control)
 
(define (make-fringe-getter tree)
(λ ()
(let loop ([tree tree])
(match tree
[(cons a d) (loop a)
(loop d)]
['() (void)]
[else (fcontrol tree)]))
(fcontrol 'done)))
 
(define (same-fringe? tree1 tree2)
(let loop ([get-fringe1 (make-fringe-getter tree1)]
[get-fringe2 (make-fringe-getter tree2)])
(% (get-fringe1)
(λ (fringe1 get-fringe1)
(% (get-fringe2)
(λ (fringe2 get-fringe2)
(and (equal? fringe1 fringe2)
(or (eq? fringe1 'done)
(loop get-fringe1 get-fringe2)))))))))
 
;; unit tests
(require rackunit)
(check-true (same-fringe? '((1 2 3) ((4 5 6) (7 8)))
'(((1 2 3) (4 5 6)) (7 8))))
(check-false (same-fringe? '((1 2 3) ((4 5 6) (7 8)))
'(((1 2 3) (4 6)) (8))))
 

[edit] REXX

[edit] Version 1 using father node

/* REXX ***************************************************************
* Same Fringe
* 1 A A
* / \ / \ / \
* / \ / \ / \
* / \ / \ / \
* 2 3 B C B C
* / \ / / \ / / \ /
* 4 5 6 D E F D E F
* / / \ / / \ / / \
* 7 8 9 G H I G * I
*
* 23.08.2012 Walter Pachl derived from
* http://rosettacode.org/wiki/Tree_traversal
* Tree A: A B D G E C F H I
* Tree B: A B D G E C F * I
**********************************************************************/

debug=0
node.=0
lvl=0
 
Call mktree 'A'
Call mktree 'B'
 
done.=0
za=root.a; leafa=node.a.za.0name
zb=root.a; leafb=node.b.zb.0name
done.a.za=1
done.b.zb=1
Do i=1 To 12
if leafa=leafb Then Do
If leafa=0 Then Do
Say 'Fringes are equal'
Leave
End
Say leafa '=' leafb
Do j=1 To 12 Until done.a.za=0
za=go_next(za,'A'); leafa=node.a.za.0name
End
done.a.za=1
Do j=1 To 12 Until done.b.zb=0
zb=go_next(zb,'B'); leafb=node.b.zb.0name
End
done.b.zb=1
End
Else Do
Select
When leafa=0 Then
Say leafb 'exceeds leaves in tree A'
When leafb=0 Then
Say leafa 'exceeds leaves in tree B'
Otherwise
Say 'First difference' leafa '<>' leafb
End
Leave
End
End
Exit
 
 
note:
/**********************************************************************
* add the node to the preorder list unless it's already there
* add the node to the level list
**********************************************************************/

Parse Arg z,t
If z<>0 &, /* it's a node */
done.z=0 Then Do /* not yet done */
wl.t=wl.t z /* add it to the preorder list*/
ll.lvl=ll.lvl z /* add it to the level list */
done.z=1 /* remember it's done */
leafl=leafl node.t.z.0name
End
Return
 
go_next: Procedure Expose node. lvl
/**********************************************************************
* find the next node to visit in the treewalk
**********************************************************************/

next=0
Parse arg z,t
If node.t.z.0left<>0 Then Do /* there is a left son */
If node.t.z.0left.done=0 Then Do /* we have not visited it */
next=node.t.z.0left /* so we go there */
node.t.z.0left.done=1 /* note we were here */
lvl=lvl+1 /* increase the level */
End
End
If next=0 Then Do /* not moved yet */
If node.t.z.0rite<>0 Then Do /* there is a right son */
If node.t.z.0rite.done=0 Then Do /* we have not visited it */
next=node.t.z.0rite /* so we go there */
node.t.z.0rite.done=1 /* note we were here */
lvl=lvl+1 /* increase the level */
End
End
End
If next=0 Then Do /* not moved yet */
next=node.t.z.0father /* go to the father */
lvl=lvl-1 /* decrease the level */
End
Return next /* that's the next node */
/* or zero if we are done */
 
mknode: Procedure Expose node.
/**********************************************************************
* create a new node
**********************************************************************/

Parse Arg name,t
z=node.t.0+1
node.t.z.0name=name
node.t.z.0father=0
node.t.z.0left =0
node.t.z.0rite =0
node.t.0=z
Return z /* number of the node just created */
 
attleft: Procedure Expose node.
/**********************************************************************
* make son the left son of father
**********************************************************************/

Parse Arg son,father,t
node.t.son.0father=father
z=node.t.father.0left
If z<>0 Then Do
node.t.z.0father=son
node.t.son.0left=z
End
node.t.father.0left=son
Return
 
attrite: Procedure Expose node.
/**********************************************************************
* make son the right son of father
**********************************************************************/

Parse Arg son,father,t
node.t.son.0father=father
z=node.t.father.0rite
If z<>0 Then Do
node.t.z.0father=son
node.t.son.0rite=z
End
node.t.father.0rite=son
le=node.t.father.0left
If le>0 Then
node.t.le.0brother=node.t.father.0rite
Return
 
mktree: Procedure Expose node. root.
/**********************************************************************
* build the tree according to the task
**********************************************************************/

Parse Arg t
If t='A' Then Do
a=mknode('A',t); root.t=a
b=mknode('B',t); Call attleft b,a,t
c=mknode('C',t); Call attrite c,a,t
d=mknode('D',t); Call attleft d,b,t
e=mknode('E',t); Call attrite e,b,t
f=mknode('F',t); Call attleft f,c,t
g=mknode('G',t); Call attleft g,d,t
h=mknode('H',t); Call attleft h,f,t
i=mknode('I',t); Call attrite i,f,t
End
Else Do
a=mknode('A',t); root.t=a
b=mknode('B',t); Call attleft b,a,t
c=mknode('C',t); Call attrite c,a,t
d=mknode('D',t); Call attleft d,b,t
e=mknode('E',t); Call attrite e,b,t
f=mknode('F',t); Call attleft f,c,t
g=mknode('G',t); Call attleft g,d,t
h=mknode('*',t); Call attleft h,f,t
i=mknode('I',t); Call attrite i,f,t
End
Return

Output:

 A = A                     
 B = B                     
 D = D                     
 G = G                     
 E = E                     
 C = C                     
 F = F                     
 First difference H <> * 

[edit] Version 2 without using father node

/* REXX ***************************************************************
* Same Fringe
= 1 A A
= / \ / \ / \
= / \ / \ / \
= / \ / \ / \
= 2 3 B C B C
= / \ / / \ / / \ /
= 4 5 6 D E F D E F
= / / \ / / \ / / \
= 7 8 9 G H I G * I
=
* 23.08.2012 Walter Pachl derived from
* http://rosettacode.org/wiki/Tree_traversal
* Tree A: A B D G E C F H I
* Tree B: A B D G E C F * I
**********************************************************************/

node.=0
 
Call mktree 'A'
Call mktree 'B'
 
sideboard.=0
 
za=root.a; leafa=node.a.za.0name
zb=root.b; leafb=node.b.zb.0name
Do i=1 To 20 Until za=0 & zb=0
If leafa=leafb Then Do
Say leafa '=' leafb
Parse Value get_next(za,'A') with za leafa
Parse Value get_next(zb,'B') with zb leafb
End
Else Do
Select
When za=0 Then Say leafb 'exceeds tree A'
When zb=0 Then Say leafa 'exceeds tree B'
Otherwise Say 'First difference' leafa '<>' leafb
End
Leave
Exit
End
End
exit
 
get_next: Procedure Expose node. sideboard.
Parse Arg za,t
Select
When node.t.za.0left<>0 Then Do
If node.t.za.0rite<>0 Then Do
z=sideboard.t.0+1
sideboard.t.z=node.t.za.0rite
sideboard.t.0=z
End
za=node.t.za.0left
End
When node.t.za.0rite<>0 Then Do
za=node.t.za.0rite
End
Otherwise Do
z=sideboard.t.0
za=sideboard.t.z
z=z-1
sideboard.t.0=z
End
End
Return za node.t.za.0name
 
mknode: Procedure Expose node.
/**********************************************************************
* create a new node
**********************************************************************/

Parse Arg name,t
z=node.t.0+1
node.t.z.0name=name
node.t.z.0father=0
node.t.z.0left =0
node.t.z.0rite =0
node.t.0=z
Return z /* number of the node just created */
 
attleft: Procedure Expose node.
/**********************************************************************
* make son the left son of father
**********************************************************************/

Parse Arg son,father,t
node.t.son.0father=father
z=node.t.father.0left
If z<>0 Then Do
node.t.z.0father=son
node.t.son.0left=z
End
node.t.father.0left=son
Return
 
attrite: Procedure Expose node.
/**********************************************************************
* make son the right son of father
**********************************************************************/

Parse Arg son,father,t
node.t.son.0father=father
z=node.t.father.0rite
If z<>0 Then Do
node.t.z.0father=son
node.t.son.0rite=z
End
node.t.father.0rite=son
le=node.t.father.0left
If le>0 Then
node.t.le.0brother=node.t.father.0rite
Return
 
mktree: Procedure Expose node. root.
/**********************************************************************
* build the tree according to the task
**********************************************************************/

Parse Arg t
If t='A' Then Do
a=mknode('A',t); root.t=a
b=mknode('B',t); Call attleft b,a,t
c=mknode('C',t); Call attrite c,a,t
d=mknode('D',t); Call attleft d,b,t
e=mknode('E',t); Call attrite e,b,t
f=mknode('F',t); Call attleft f,c,t
g=mknode('G',t); Call attleft g,d,t
h=mknode('H',t); Call attleft h,f,t
i=mknode('I',t); Call attrite i,f,t
End
Else Do
a=mknode('A',t); root.t=a
b=mknode('B',t); Call attleft b,a,t
c=mknode('C',t); Call attrite c,a,t
d=mknode('D',t); Call attleft d,b,t
e=mknode('E',t); Call attrite e,b,t
f=mknode('F',t); Call attleft f,c,t
g=mknode('G',t); Call attleft g,d,t
h=mknode('*',t); Call attleft h,f,t
i=mknode('I',t); Call attrite i,f,t
End
Return

Output is the same as for Version 1

[edit] version 1.1

This REXX example is a re-written program that mimics the first version (above).

This version has:

  • elided a subroutine
  • elided superfluous   do ── end   groups
  • elided some stemmed array tails
  • elided some REXX variables   (lvl, debug, ···)
  • simplified some stem names
  • displays the tree   (as an ASCII display)
  • changed tree names so as to not conflict with leaf names
  • uses non-case sensitive tree names
  • used boolean based variables as logicals
  • expanded message texts
  • combined subroutines   attleft   and   attright   into one
  • streamlined the   make_tree   subroutine
/*REXX program examines the leaves of two binary trees (as shown below).*/
_=left('',28); say _ " A A "
say _ " / \ ◄════1st tree / \ "
say _ " / \ / \ "
say _ " / \ / \ "
say _ " B C B C "
say _ " / \ / 2nd tree════► / \ / "
say _ " D E F D E F "
say _ " / / \ / / \ "
say _ "G H I G δ I "
say
#=0; done.=0; node.=0 /*initialize # leaves,DONE.,NODE.*/
call make_tree 1 /*define tree number 1 (1st tree)*/
call make_tree 2 /* " " " 2 (2nd " )*/
z1=root.1; L1=node.1.z1; done.1.z1=1 /*L1 is a leaf on tree number 1. */
z2=z1; L2=node.2.z2; done.2.z2=1 /*L2 " " " " " " 2. */
 
do # % 2 /*loop for the number of leaves. */
if L1==L2 then do
if L1==0 then call sayX 'The trees are equal.'
say ' The ' L1 " leaf is identical in both trees."
do until \done.1.z1
z1=go_next(z1,1); L1=node.1.z1
end
done.1.z1=1
do until \done.2.z2
z2=go_next(z2,2); L2=node.2.z2
end
done.2.z2=1
end
else select
when L1==0 then call sayX L2 'exceeds leaves in 1st tree'
when L2==0 then call sayX L1 'exceeds leaves in 2nd tree'
otherwise call sayX 'A difference is: ' L1 '¬=' L2
end /*select*/
end /*# % 2*/
exit
/*──────────────────────────────────GO_NEXT subroutine──────────────────*/
go_next: procedure expose node.; arg q,t /*find next node.*/
next=.
if node.t.q._Lson\==0 &, /*is there a left branch in tree?*/
node.t.q._Lson.vis==0 then do /*has this node been visited yet?*/
next=node.t.q._Lson /*──► next node. */
node.t.q._Lson.vis=1 /*leftside done. */
end
if next==. &,
node.t.q._Rson\==0 &, /*is there a right tree branch ? */
node.t.q._Rson.vis==0 then do /*has this node been VISited yet?*/
next=node.t.q._Rson /*──► next node. */
node.t.q._Rson.vis=1 /*rightside done.*/
end
if next==. then next=node.t.q._dad /*process the father node. */
return next /*the next node (or 0, if done).*/
/*──────────────────────────────────MAKE_NODE subroutine────────────────*/
make_node: parse arg name,t; # = #+1 /*make a new node/branch on tree.*/
q = node.t.0 + 1  ; node.t.q = name  ; node.t.q._dad = 0
node.t.q._Lson = 0 ; node.t.q._Rson = 0  ; node.t.0 = q
return q /*number of the node just created*/
/*──────────────────────────────────MAKE_TREE subroutine────────────────*/
make_tree: procedure expose node. root. #; parse arg tree /*build trees*/
if tree==1 then hhh='H' /* [↓] must be a wood duck*/
else hhh='δ' /*the odd duck in the whole tree.*/
a=make_node('A', tree); root.tree=a
b=make_node('B', tree); call son 'L', b,a,tree
c=make_node('C', tree); call son 'R', c,a,tree
d=make_node('D', tree); call son 'L', d,b,tree
e=make_node('E', tree); call son 'R', e,b,tree
f=make_node('F', tree); call son 'L', f,c,tree
g=make_node('G', tree); call son 'L', g,d,tree
/*quacks like a duck?*/ h=make_node(hhh, tree); call son 'L', h,f,tree
i=make_node('I', tree); call son 'R', i,f,tree
return
/*──────────────────────────────────SAYX subroutine─────────────────────*/
sayX: say; say arg(1); say; exit /*tell msg and exit.*/
/*──────────────────────────────────SON subroutine──────────────────────*/
son: procedure expose node.; parse arg ?,son,dad,t; LR = '_'?"SON"
node.t.son._dad=dad; q=node.t.dad.LR /*define which son [↑] */
if q\==0 then do; node.t.q._dad=son; node.t.son.LR=q; end
node.t.dad.LR=son
if ?=='R' & node.t.dad.LR>0 then node.t.le._brother=node.t.dad.LR
return

output

                                     A                                   A
                                    / \    ◄────1st tree                / \
                                   /   \                               /   \
                                  /     \                             /     \
                                 B       C                           B       C
                                / \     /          2nd tree────►    / \     /
                               D   E   F                           D   E   F
                              /       / \                         /       / \
                             G       H   I                       G       δ   I

    The  A  leaf is identical in both trees.
    The  B  leaf is identical in both trees.
    The  D  leaf is identical in both trees.
    The  G  leaf is identical in both trees.
    The  E  leaf is identical in both trees.
    The  C  leaf is identical in both trees.
    The  F  leaf is identical in both trees.

A difference is:  H ¬= δ

[edit] Scheme

Descend provides a list, or stack, of the leftmost unvisited nodes at each level of the tree. Two such lists are used as cursors to keep track of the remaining nodes. The main loop compares the top of each list (ie the leftmost remaining node) and breaks with false if different, or calls Ascend to update the lists. Updating may require calling Descend again if more unvisited left-nodes are found. If the end of both lists is reached simultaneously, and therefore the end of both trees, true is returned.

; binary tree helpers from "Structure and Interpretation of Computer Programs" 2.3.3
(define (entry tree) (car tree))
(define (left-branch tree) (cadr tree))
(define (right-branch tree) (caddr tree))
(define (make-tree entry left right)
(list entry left right))
 
; returns a list of leftmost nodes from each level of the tree
(define (descend tree ls)
(if (null? (left-branch tree))
(cons tree ls)
(descend (left-branch tree) (cons tree ls))))
 
; updates the list to contain leftmost nodes from each remaining level
(define (ascend ls)
(cond
((and (null? (cdr ls)) (null? (right-branch (car ls)))) '())
((null? (right-branch (car ls))) (cdr ls))
(else
(let ((ls (cons (right-branch (car ls))
(cdr ls))))
(if (null? (left-branch (car ls)))
ls
(descend (left-branch (car ls)) ls))))))
 
; loops thru each list until the end (true) or nodes are unequal (false)
(define (same-fringe? t1 t2)
(let next ((l1 (descend t1 '()))
(l2 (descend t2 '())))
(cond
((and (null? l1) (null? l2)) #t)
((or (null? l1)
(null? l2)
(not (eq? (entry (car l1)) (entry (car l2))))) #f)
(else (next (ascend l1) (ascend l2))))))
Output:
> (same-fringe? (list 1 '() (list 2 '() (list 3 '() '()))) (list 3 (list 2 (list 1 '() '()) '()) '()))
#t

[edit] Tcl

Works with: Tcl version 8.6
Library: Tcllib (Package: struct::tree)
package require Tcl 8.6
package require struct::tree
 
# A wrapper round a coroutine for iterating over the leaves of a tree in order
proc leafiterator {tree} {
coroutine coro[incr ::coroutines] apply {tree {
yield [info coroutine]
$tree walk [$tree rootname] node {
if {[$tree isleaf $node]} {
yield $node
}
}
yieldto break
}} $tree
}
 
# Compare two trees for equality of their leaf node names
proc samefringe {tree1 tree2} {
set c1 [leafiterator $tree1]
set c2 [leafiterator $tree2]
try {
while 1 {
if {[set l1 [$c1]] ne [set l2 [$c2]]} {
puts "$l1 != $l2"; # Just so we can see where we failed
return 0
}
}
return 1
} finally {
rename $c1 {}
rename $c2 {}
}
}

Demonstrating:

# Make some trees to compare...
struct::tree t1 deserialize {
root {} {}
a 0 {}
d 3 {}
e 3 {}
b 0 {}
c 0 {}
}
struct::tree t2 deserialize {
root {} {}
a 0 {}
d 3 {}
e 3 {}
b 0 {}
cc 0 {}
}
 
# Print the boolean result of doing the comparison
puts [samefringe t1 t2]
Output:
c != cc
0

[edit] zkl

Translation of: Icon and Unicon
var G=Utils.Generator;
//Tree: (node,left,right) or (leaf) or (node,left) ...
aTree := T(1, T(2, T(4, T(7)), T(5)), T(3, T(6, T(8), T(9))));
bTree := aTree;
println("aTree and bTree ",sameFringe(aTree,bTree) and "have" or "don't have",
" the same leaves.");
cTree := T(1, T(2, T(4, T(7)), T(5)), T(3, T(6, T(8))));
dTree := T(1, T(2, T(4, T(7)), T(5)), T(3, T(6, T(8), T(9))));
println("cTree and dTree ",sameFringe(cTree,dTree) and "have"or"don't have",
" the same leaves.");
 
fcn sameFringe(a,b){ same(G(genLeaves,a),G(genLeaves,b)) }
 
fcn same(g1,g2){ //(Generator,Generator)
foreach n1,n2 in (g1.zip(g2)){ //-->(int,int) ...
if(n1 != n2) return(); // == return(Void)
}
return(not (g2._next() or g2._next())); //-->False if g1 or g2 has leaves
}
 
fcn genLeaves(tree){
switch(tree.len()){ // (), (leaf), (node,left, [right])
case(1){ vm.yield(tree[0]) } // leaf: int
case(2){ genLeaves(tree[1]); }
else { genLeaves(tree[1]); genLeaves(tree[2]); }
}
}
Output:
aTree and bTree have the same leaves.
cTree and dTree don't have the same leaves.
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