Tree traversal
From Rosetta Code
Tree traversal is a programming task. Visitors like you are encouraged to solve it according to the task description, using any language they may happen to know.
Implement a binary tree where each node carries an integer, and implement preoder, inorder, postorder and level-order traversal. Use those traversals to output the following tree:
1
/ \
/ \
/ \
2 3
/ \ /
4 5 6
/ / \
7 8 9
The correct output should look like this:
preorder: 1 2 4 7 5 3 6 8 9 inorder: 7 4 2 5 1 8 6 9 3 postorder: 7 4 5 2 8 9 6 3 1 level-order: 1 2 3 4 5 6 7 8 9
This article has more information on traversing trees.
Contents |
[edit] Ada
with Ada.Text_Io; use Ada.Text_Io;
with Ada.Unchecked_Deallocation;
with Ada.Containers.Doubly_Linked_Lists;
procedure Tree_Traversal is
type Node;
type Node_Access is access Node;
type Node is record
Left : Node_Access := null;
Right : Node_Access := null;
Data : Integer;
end record;
procedure Destroy_Tree(N : in out Node_Access) is
procedure free is new Ada.Unchecked_Deallocation(Node, Node_Access);
begin
if N.Left /= null then
Destroy_Tree(N.Left);
end if;
if N.Right /= null then
Destroy_Tree(N.Right);
end if;
Free(N);
end Destroy_Tree;
function Tree(Value : Integer; Left : Node_Access; Right : Node_Access) return Node_Access is
Temp : Node_Access := new Node;
begin
Temp.Data := Value;
Temp.Left := Left;
Temp.Right := Right;
return Temp;
end Tree;
procedure Preorder(N : Node_Access) is
begin
Put(Integer'Image(N.Data));
if N.Left /= null then
Preorder(N.Left);
end if;
if N.Right /= null then
Preorder(N.Right);
end if;
end Preorder;
procedure Inorder(N : Node_Access) is
begin
if N.Left /= null then
Inorder(N.Left);
end if;
Put(Integer'Image(N.Data));
if N.Right /= null then
Inorder(N.Right);
end if;
end Inorder;
procedure Postorder(N : Node_Access) is
begin
if N.Left /= null then
Postorder(N.Left);
end if;
if N.Right /= null then
Postorder(N.Right);
end if;
Put(Integer'Image(N.Data));
end Postorder;
procedure Levelorder(N : Node_Access) is
package Queues is new Ada.Containers.Doubly_Linked_Lists(Node_Access);
use Queues;
Node_Queue : List;
Next : Node_Access;
begin
Node_Queue.Append(N);
while not Is_Empty(Node_Queue) loop
Next := First_Element(Node_Queue);
Delete_First(Node_Queue);
Put(Integer'Image(Next.Data));
if Next.Left /= null then
Node_Queue.Append(Next.Left);
end if;
if Next.Right /= null then
Node_Queue.Append(Next.Right);
end if;
end loop;
end Levelorder;
N : Node_Access;
begin
N := Tree(1,
Tree(2,
Tree(4,
Tree(7, null, null),
null),
Tree(5, null, null)),
Tree(3,
Tree(6,
Tree(8, null, null),
Tree(9, null, null)),
null));
Put("preorder: ");
Preorder(N);
New_Line;
Put("inorder: ");
Inorder(N);
New_Line;
Put("postorder: ");
Postorder(N);
New_Line;
Put("level order: ");
Levelorder(N);
New_Line;
Destroy_Tree(N);
end Tree_traversal;
[edit] C
#include <stdlib.h>
#include <stdio.h>
typedef struct node_s
{
int value;
struct node_s* left;
struct node_s* right;
} *node;
node tree(int v, node l, node r)
{
node n = malloc(sizeof(struct node_s));
n->value = v;
n->left = l;
n->right = r;
return n;
}
void destroy_tree(node n)
{
if (n->left)
destroy_tree(n->left);
if (n->right)
destroy_tree(n->right);
free(n);
}
void preorder(node n, void (*f)(int))
{
f(n->value);
if (n->left)
preorder(n->left, f);
if (n->right)
preorder(n->right, f);
}
void inorder(node n, void (*f)(int))
{
if (n->left)
inorder(n->left, f);
f(n->value);
if (n->right)
inorder(n->right, f);
}
void postorder(node n, void (*f)(int))
{
if (n->left)
postorder(n->left, f);
if (n->right)
postorder(n->right, f);
f(n->value);
}
/* helper queue for levelorder */
typedef struct qnode_s
{
struct qnode_s* next;
node value;
} *qnode;
typedef struct { qnode begin, end; } queue;
void enqueue(queue* q, node n)
{
qnode node = malloc(sizeof(struct qnode_s));
node->value = n;
node->next = 0;
if (q->end)
q->end->next = node;
else
q->begin = node;
q->end = node;
}
node dequeue(queue* q)
{
node tmp = q->begin->value;
qnode second = q->begin->next;
free(q->begin);
q->begin = second;
if (!q->begin)
q->end = 0;
return tmp;
}
int queue_empty(queue* q)
{
return !q->begin;
}
void levelorder(node n, void(*f)(int))
{
queue nodequeue = {};
enqueue(&nodequeue, n);
while (!queue_empty(&nodequeue))
{
node next = dequeue(&nodequeue);
f(next->value);
if (next->left)
enqueue(&nodequeue, next->left);
if (next->right)
enqueue(&nodequeue, next->right);
}
}
void print(int n)
{
printf("%d ", n);
}
int main()
{
node n = tree(1,
tree(2,
tree(4,
tree(7, 0, 0),
0),
tree(5, 0, 0)),
tree(3,
tree(6,
tree(8, 0, 0),
tree(9, 0, 0)),
0));
printf("preorder: ");
preorder(n, print);
printf("\n");
printf("inorder: ");
inorder(n, print);
printf("\n");
printf("postorder: ");
postorder(n, print);
printf("\n");
printf("level-order: ");
levelorder(n, print);
printf("\n");
destroy_tree(n);
return 0;
}
[edit] C++
Compiler: g++ (version 4.3.2 20081105 (Red Hat 4.3.2-7))
Library: Boost
#include <boost/scoped_ptr.hpp>
#include <iostream>
#include <queue>
template<typename T>
class TreeNode {
public:
TreeNode(const T& n, TreeNode* left = NULL, TreeNode* right = NULL)
: mValue(n),
mLeft(left),
mRight(right) {}
T getValue() const {
return mValue;
}
TreeNode* left() const {
return mLeft.get();
}
TreeNode* right() const {
return mRight.get();
}
void preorderTraverse() const {
std::cout << " " << getValue();
if(mLeft) { mLeft->preorderTraverse(); }
if(mRight) { mRight->preorderTraverse(); }
}
void inorderTraverse() const {
if(mLeft) { mLeft->inorderTraverse(); }
std::cout << " " << getValue();
if(mRight) { mRight->inorderTraverse(); }
}
void postorderTraverse() const {
if(mLeft) { mLeft->postorderTraverse(); }
if(mRight) { mRight->postorderTraverse(); }
std::cout << " " << getValue();
}
void levelorderTraverse() const {
std::queue<const TreeNode*> q;
q.push(this);
while(!q.empty()) {
const TreeNode* n = q.front();
q.pop();
std::cout << " " << n->getValue();
if(n->left()) { q.push(n->left()); }
if(n->right()) { q.push(n->right()); }
}
}
protected:
T mValue;
boost::scoped_ptr<TreeNode> mLeft;
boost::scoped_ptr<TreeNode> mRight;
private:
TreeNode();
};
int main() {
TreeNode<int> root(1,
new TreeNode<int>(2,
new TreeNode<int>(4,
new TreeNode<int>(7)),
new TreeNode<int>(5)),
new TreeNode<int>(3,
new TreeNode<int>(6,
new TreeNode<int>(8),
new TreeNode<int>(9))));
std::cout << "preorder: ";
root.preorderTraverse();
std::cout << std::endl;
std::cout << "inorder: ";
root.inorderTraverse();
std::cout << std::endl;
std::cout << "postorder: ";
root.postorderTraverse();
std::cout << std::endl;
std::cout << "level-order:";
root.levelorderTraverse();
std::cout << std::endl;
return 0;
}
[edit] Clojure
(defn walk [node f order]
(when node
(doseq [o order]
(if (= o :visit)
(f (:val node))
(walk (node o) f order)))))
(defn preorder [node f]
(walk node f [:visit :left :right]))
(defn inorder [node f]
(walk node f [:left :visit :right]))
(defn postorder [node f]
(walk node f [:left :right :visit]))
(defn queue [& xs]
(when (seq xs)
(apply conj clojure.lang.PersistentQueue/EMPTY xs)))
(defn level-order [root f]
(loop [q (queue root)]
(when-not (empty? q)
(if-let [node (first q)]
(do
(f (:val node))
(recur (conj (pop q) (:left node) (:right node))))
(recur (pop q))))))
(defn vec-to-tree [t]
(if (vector? t)
(let [[val left right] t]
{:val val
:left (vec-to-tree left)
:right (vec-to-tree right)})
t))
(let [tree (vec-to-tree [1 [2 [4 [7]] [5]] [3 [6 [8] [9]]]])
fs '[preorder inorder postorder level-order]
pr-node #(print (format "%2d" %))]
(doseq [f fs]
(print (format "%-12s" (str f ":")))
((resolve f) tree pr-node)
(println)))
[edit] Common Lisp
(defun preorder (node f)
(when node
(funcall f (first node))
(preorder (second node) f)
(preorder (third node) f)))
(defun inorder (node f)
(when node
(inorder (second node) f)
(funcall f (first node))
(inorder (third node) f)))
(defun postorder (node f)
(when node
(postorder (second node) f)
(funcall f (first node))
(postorder (third node) f)))
(defun level-order (node f)
(loop with level = (list node)
while level
do
(setf level (loop for node in level
when node
do (funcall f (first node))
and collect (second node)
and collect (third node)))))
(defparameter *tree* '(1 (2 (4 (7))
(5))
(3 (6 (8)
(9)))))
(defun show (traversal-function)
(format t "~&~(~A~):~12,0T" traversal-function)
(funcall traversal-function *tree* (lambda (value) (format t " ~A" value))))
(map nil #'show '(preorder inorder postorder level-order))
Output:
preorder: 1 2 4 7 5 3 6 8 9 inorder: 7 4 2 5 1 8 6 9 3 postorder: 7 4 2 5 1 8 6 9 3 level-order: 1 2 3 4 5 6 7 8 9
[edit] D
Code for D V.2. This code is very generic, if you need it less generic it can be shortened.
import std.stdio: write, writeln;
class Node(T) {
T data;
Node left, right;
this(T data, Node left=null, Node right=null) {
this.data = data;
this.left = left;
this.right = right;
}
}
// static templated opCall can't be used in Node
auto node(T)(T data, Node!T left=null, Node!T right=null) {
return new Node!T(data, left, right);
}
void show(T)(T x) {
write(x, " ");
}
enum Visit { pre, inv, post }
// visitor can be any kind of callable or it uses a default visitor.
// TNode can be any kind of Node, with data, left and right fields,
// so this is more generic than a member function of Node.
void backtrackingOrder(Visit v, TNode, TyF=void*)(TNode node, TyF visitor=null) {
static if (is(TyF == void*)) auto truevisitor = &show!(typeof(node.data));
else auto truevisitor = visitor;
if (node !is null) {
static if (v == Visit.pre) truevisitor(node.data);
backtrackingOrder!v(node.left, visitor);
static if (v == Visit.inv) truevisitor(node.data);
backtrackingOrder!v(node.right, visitor);
static if (v == Visit.post) truevisitor(node.data);
}
}
void levelOrder(TNode, TyF=void*)(TNode node, TyF visitor=null, TNode[] more=[]) {
static if (is(TyF == void*)) auto truevisitor = &show!(typeof(node.data));
else auto truevisitor = visitor;
if (node !is null) {
more ~= [node.left, node.right];
truevisitor(node.data);
}
if (more.length)
levelOrder(more[0], truevisitor, more[1..$]);
}
void main() {
auto tree = node(1,
node(2,
node(4,
node(7)),
node(5)),
node(3,
node(6,
node(8),
node(9))));
write(" preOrder: ");
backtrackingOrder!(Visit.pre)(tree);
write("\n inorder: ");
backtrackingOrder!(Visit.inv)(tree);
write("\n postOrder: ");
backtrackingOrder!(Visit.post)(tree);
write("\nlevelorder: ");
levelOrder(tree);
writeln();
}
Output:
preOrder: 1 2 4 7 5 3 6 8 9 inOrder: 7 4 2 5 1 8 6 9 3 postOrder: 7 4 5 2 8 9 6 3 1 levelorder: 1 2 3 4 5 6 7 8 9
[edit] E
def btree := [1, [2, [4, [7, null, null],
null],
[5, null, null]],
[3, [6, [8, null, null],
[9, null, null]],
null]]
def backtrackingOrder(node, pre, mid, post) {
switch (node) {
match ==null {}
match [value, left, right] {
pre(value)
backtrackingOrder(left, pre, mid, post)
mid(value)
backtrackingOrder(right, pre, mid, post)
post(value)
}
}
}
def levelOrder(root, func) {
var level := [root].diverge()
while (level.size() > 0) {
for node in level.removeRun(0) {
switch (node) {
match ==null {}
match [value, left, right] {
func(value)
level.push(left)
level.push(right)
} } } } }
print("preorder: ")
backtrackingOrder(btree, fn v { print(" ", v) }, fn _ {}, fn _ {})
println()
print("inorder: ")
backtrackingOrder(btree, fn _ {}, fn v { print(" ", v) }, fn _ {})
println()
print("postorder: ")
backtrackingOrder(btree, fn _ {}, fn _ {}, fn v { print(" ", v) })
println()
print("level-order:")
levelOrder(btree, fn v { print(" ", v) })
println()
[edit] Erlang
-module(tree_traversal).
-export([main/0]).
-export([preorder/2, inorder/2, postorder/2, levelorder/2]).
-export([tnode/0, tnode/1, tnode/3]).
-define(NEWLINE, io:format("~n")).
tnode() -> {}.
tnode(V) -> {node, V, {}, {}}.
tnode(V,L,R) -> {node, V, L, R}.
preorder(_,{}) -> ok;
preorder(F,{node,V,L,R}) ->
F(V), preorder(F,L), preorder(F,R).
inorder(_,{}) -> ok;
inorder(F,{node,V,L,R}) ->
inorder(F,L), F(V), inorder(F,R).
postorder(_,{}) -> ok;
postorder(F,{node,V,L,R}) ->
postorder(F,L), postorder(F,R), F(V).
levelorder(_, []) -> [];
levelorder(F, [{}|T]) -> levelorder(F, T);
levelorder(F, [{node,V,L,R}|T]) ->
F(V), levelorder(F, T++[L,R]);
levelorder(F, X) -> levelorder(F, [X]).
main() ->
Tree = tnode(1,
tnode(2,
tnode(4, tnode(7), tnode()),
tnode(5, tnode(), tnode())),
tnode(3,
tnode(6, tnode(8), tnode(9)),
tnode())),
F = fun(X) -> io:format("~p ",[X]) end,
preorder(F, Tree), ?NEWLINE,
inorder(F, Tree), ?NEWLINE,
postorder(F, Tree), ?NEWLINE,
levelorder(F, Tree), ?NEWLINE.
Output:
1 2 4 7 5 3 6 8 9 7 4 2 5 1 8 6 9 3 7 4 5 2 8 9 6 3 1 1 2 3 4 5 6 7 8 9
[edit] Factor
USING: accessors combinators deques dlists fry io kernel
math.parser ;
IN: rosetta.tree-traversal
TUPLE: node data left right ;
CONSTANT: example-tree
T{ node f 1
T{ node f 2
T{ node f 4
T{ node f 7 f f }
f
}
T{ node f 5 f f }
}
T{ node f 3
T{ node f 6
T{ node f 8 f f }
T{ node f 9 f f }
}
f
}
}
: preorder ( node quot: ( data -- ) -- )
[ [ data>> ] dip call ]
[ [ left>> ] dip over [ preorder ] [ 2drop ] if ]
[ [ right>> ] dip over [ preorder ] [ 2drop ] if ]
2tri ; inline recursive
: inorder ( node quot: ( data -- ) -- )
[ [ left>> ] dip over [ inorder ] [ 2drop ] if ]
[ [ data>> ] dip call ]
[ [ right>> ] dip over [ inorder ] [ 2drop ] if ]
2tri ; inline recursive
: postorder ( node quot: ( data -- ) -- )
[ [ left>> ] dip over [ postorder ] [ 2drop ] if ]
[ [ right>> ] dip over [ postorder ] [ 2drop ] if ]
[ [ data>> ] dip call ]
2tri ; inline recursive
: (levelorder) ( dlist quot: ( data -- ) -- )
over deque-empty? [ 2drop ] [
[ dup pop-front ] dip {
[ [ data>> ] dip call drop ]
[ drop left>> [ swap push-back ] [ drop ] if* ]
[ drop right>> [ swap push-back ] [ drop ] if* ]
[ nip (levelorder) ]
} 3cleave
] if ; inline recursive
: levelorder ( node quot: ( data -- ) -- )
[ 1dlist ] dip (levelorder) ; inline
: levelorder2 ( node quot: ( data -- ) -- )
[ 1dlist ] dip
[ dup deque-empty? not ] swap '[
dup pop-front
[ data>> @ ]
[ left>> [ over push-back ] when* ]
[ right>> [ over push-back ] when* ] tri
] while drop ; inline
: main ( -- )
example-tree [ number>string write " " write ] {
[ "preorder: " write preorder nl ]
[ "inorder: " write inorder nl ]
[ "postorder: " write postorder nl ]
[ "levelorder: " write levelorder nl ]
[ "levelorder2: " write levelorder2 nl ]
} 2cleave ;
[edit] Forth
\ binary tree (dictionary)
: node ( l r data -- node ) here >r , , , r> ;
: leaf ( data -- node ) 0 0 rot node ;
: >data ( node -- ) @ ;
: >right ( node -- ) cell+ @ ;
: >left ( node -- ) cell+ cell+ @ ;
: preorder ( xt tree -- )
dup 0= if 2drop exit then
2dup >data swap execute
2dup >left recurse
>right recurse ;
: inorder ( xt tree -- )
dup 0= if 2drop exit then
2dup >left recurse
2dup >data swap execute
>right recurse ;
: postorder ( xt tree -- )
dup 0= if 2drop exit then
2dup >left recurse
2dup >right recurse
>data swap execute ;
: max-depth ( tree -- n )
dup 0= if exit then
dup >left recurse
swap >right recurse max 1+ ;
defer depthaction
: depthorder ( depth tree -- )
dup 0= if 2drop exit then
over 0=
if >data depthaction drop
else over 1- over >left recurse
swap 1- swap >right recurse
then ;
: levelorder ( xt tree -- )
swap is depthaction
dup max-depth 0 ?do
i over depthorder
loop drop ;
7 leaf 0 4 node
5 leaf 2 node
8 leaf 9 leaf 6 node
0 3 node 1 node value tree
cr ' . tree preorder \ 1 2 4 7 5 3 6 8 9
cr ' . tree inorder \ 7 4 2 5 1 8 6 9 3
cr ' . tree postorder \ 7 4 5 2 8 9 6 3 1
cr tree max-depth . \ 4
cr ' . tree levelorder \ 1 2 3 4 5 6 7 8 9
[edit] Haskell
data Tree a = Empty
| Node { value :: a,
left :: Tree a,
right :: Tree a }
preorder, inorder, postorder, levelorder :: Tree a -> [a]
preorder Empty = []
preorder (Node v l r) = [v]
++ preorder l
++ preorder r
inorder Empty = []
inorder (Node v l r) = inorder l
++ [v]
++ inorder r
postorder Empty = []
postorder (Node v l r) = postorder l
++ postorder r
++ [v]
levelorder x = loop [x]
where loop [] = []
loop (Empty : xs) = loop xs
loop (Node v l r : xs) = v : loop (xs ++ [l,r])
tree :: Tree Int
tree = Node 1
(Node 2
(Node 4
(Node 7 Empty Empty)
Empty)
(Node 5 Empty Empty))
(Node 3
(Node 6
(Node 8 Empty Empty)
(Node 9 Empty Empty))
Empty)
main :: IO ()
main = do print $ preorder tree
print $ inorder tree
print $ postorder tree
print $ levelorder tree
Output:
[1,2,4,7,5,3,6,8,9] [7,4,2,5,1,8,6,9,3] [7,4,5,2,8,9,6,3,1] [1,2,3,4,5,6,7,8,9]
[edit] J
preorder=: ]S:0
postorder=: ([:; postorder&.>@}.) , >@{.
levelorder=: ;@({::L:1 _~ [: (/: #@>) <S:1@{::)
inorder=: ([:; inorder&.>@(''"_`(1&{)@.(1<#))) , >@{. , [:; inorder&.>@}.@}.
Required example:
N2=: conjunction def '(<m),(<n),<y'
N1=: conjunction def '(<m),<n'
L=: adverb def '<m'
N2=: conjunction def '(<m),(<n),<y'
N1=: adverb def '(<m),<y'
L=: adverb def '<m'
tree=: 1 N2 (2 N2 (4 N1 (7 L)) 5 L) 3 N1 6 N2 (8 L) 9 L
This tree is organized in a pre-order fashion
preorder tree 1 2 4 7 5 3 6 8 9
post-order is not that much different from pre-order, except that the children must extracted before the parent.
postorder tree 7 4 5 2 9 8 6 3 1
Implementing in-order is more complex because we must sometimes test whether we have any leaves, instead of relying on J's implicit looping over lists
inorder tree 7 4 2 5 1 9 8 6 3
level-order can be accomplished by constructing a map of the locations of the leaves, sorting these map locations by their non-leaf indices and using the result to extract all leaves from the tree. Elements at the same level with the same parent will have the same sort keys and thus be extracted in preorder fashion, which works just fine.
levelorder tree 1 2 3 4 5 6 7 8 9
For J novices, here's the tree instance with a few redundant parenthesis:
tree=: 1 N2 (2 N2 (4 N1 (7 L)) (5 L)) (3 N1 (6 N2 (8 L) (9 L)))
Syntactically, N2 is a binary node expressed as m N2 n y. N1 is a node with a single child, expressed as m N2 y. L is a leaf node, expressed as m L. (And n must be parenthesized if it is not a single word.)
[edit] Java
Works with: Java version 1.5+
import java.util.Queue;
import java.util.LinkedList;
public class TreeTraverse {
private static class Node<T>{
public Node<T> left;
public Node<T> right;
public T data;
public Node(T data){
this.data = data;
}
public Node<T> getLeft() {
return left;
}
public void setLeft(Node<T> left) {
this.left = left;
}
public Node<T> getRight() {
return right;
}
public void setRight(Node<T> right) {
this.right = right;
}
}
public static void preorder(Node<?> n){
if (n != null) {
System.out.print(n.data + " ");
preorder(n.getLeft());
preorder(n.getRight());
}
}
public static void inorder(Node<?> n){
if (n != null) {
inorder(n.getLeft());
System.out.print(n.data + " ");
inorder(n.getRight());
}
}
public static void postorder(Node<?> n){
if (n != null){
postorder(n.getLeft());
postorder(n.getRight());
System.out.print(n.data + " ");
}
}
public static void levelorder(Node<?> n) {
Queue<Node<?>> nodequeue = new LinkedList<Node<?>>();
if (n != null)
nodequeue.add(n);
while (!nodequeue.isEmpty()) {
Node<?> next = nodequeue.remove();
System.out.print(next.data + " ");
if (next.getLeft() != null) {
nodequeue.add(next.getLeft());
}
if (next.getRight() != null) {
nodequeue.add(next.getRight());
}
}
}
public static void main(String[] args){
Node<Integer> one = new Node<Integer>(1);
Node<Integer> two = new Node<Integer>(2);
Node<Integer> three = new Node<Integer>(3);
Node<Integer> four = new Node<Integer>(4);
Node<Integer> five = new Node<Integer>(5);
Node<Integer> six = new Node<Integer>(6);
Node<Integer> seven = new Node<Integer>(7);
Node<Integer> eight = new Node<Integer>(8);
Node<Integer> nine = new Node<Integer>(9);
one.setLeft(two);
one.setRight(three);
two.setLeft(four);
two.setRight(five);
three.setLeft(six);
four.setLeft(seven);
six.setLeft(eight);
six.setRight(nine);
preorder(one);
System.out.println();
inorder(one);
System.out.println();
postorder(one);
System.out.println();
levelorder(one);
System.out.println();
}
}
Output:
1 2 4 7 5 3 6 8 9 7 4 2 5 1 8 6 9 3 7 4 5 2 8 9 6 3 1 1 2 3 4 5 6 7 8 9
[edit] JavaScript
inspired by Ruby
function BinaryTree(value, left, right) {
this.value = value;
this.left = left;
this.right = right;
}
BinaryTree.prototype.preorder = function(f) {this.walk(f,['this','left','right'])}
BinaryTree.prototype.inorder = function(f) {this.walk(f,['left','this','right'])}
BinaryTree.prototype.postorder = function(f) {this.walk(f,['left','right','this'])}
BinaryTree.prototype.walk = function(func, order) {
for (var i in order)
switch (order[i]) {
case "this": func(this.value); break;
case "left": if (this.left) this.left.walk(func, order); break;
case "right": if (this.right) this.right.walk(func, order); break;
}
}
BinaryTree.prototype.levelorder = function(func) {
var queue = [this];
while (queue.length != 0) {
var node = queue.shift();
func(node.value);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
}
// convenience function for creating a binary tree
function createBinaryTreeFromArray(ary) {
var left = null, right = null;
if (ary[1]) left = createBinaryTreeFromArray(ary[1]);
if (ary[2]) right = createBinaryTreeFromArray(ary[2]);
return new BinaryTree(ary[0], left, right);
}
var tree = createBinaryTreeFromArray([1, [2, [4, [7]], [5]], [3, [6, [8],[9]]]]);
print("*** preorder ***"); tree.preorder(print);
print("*** inorder ***"); tree.inorder(print);
print("*** postorder ***"); tree.postorder(print);
print("*** levelorder ***"); tree.levelorder(print);
[edit] Logo
; nodes are [data left right], use "first" to get data
to node.left :node
if empty? butfirst :node [output []]
output first butfirst :node
end
to node.right :node
if empty? butfirst :node [output []]
if empty? butfirst butfirst :node [output []]
output first butfirst butfirst :node
end
to max :a :b
output ifelse :a > :b [:a] [:b]
end
to tree.depth :tree
if empty? :tree [output 0]
output 1 + max tree.depth node.left :tree tree.depth node.right :tree
end
to pre.order :tree :action
if empty? :tree [stop]
invoke :action first :tree
pre.order node.left :tree :action
pre.order node.right :tree :action
end
to in.order :tree :action
if empty? :tree [stop]
in.order node.left :tree :action
invoke :action first :tree
in.order node.right :tree :action
end
to post.order :tree :action
if empty? :tree [stop]
post.order node.left :tree :action
post.order node.right :tree :action
invoke :action first :tree
end
to at.depth :n :tree :action
if empty? :tree [stop]
ifelse :n = 1 [invoke :action first :tree] [
at.depth :n-1 node.left :tree :action
at.depth :n-1 node.right :tree :action
]
end
to level.order :tree :action
for [i 1 [tree.depth :tree]] [at.depth :i :tree :action]
end
make "tree [1 [2 [4 [7]]
[5]]
[3 [6 [8]
[9]]]]
pre.order :tree [(type ? "| |)] (print)
in.order :tree [(type ? "| |)] (print)
post.order :tree [(type ? "| |)] (print)
level.order :tree [(type ? "| |)] (print)
[edit] OCaml
type 'a tree = Empty
| Node of 'a * 'a tree * 'a tree
let rec preorder f = function
Empty -> ()
| Node (v,l,r) -> f v;
preorder f l;
preorder f r
let rec inorder f = function
Empty -> ()
| Node (v,l,r) -> inorder f l;
f v;
inorder f r
let rec postorder f = function
Empty -> ()
| Node (v,l,r) -> preorder f l;
preorder f r;
f v
let levelorder f x =
let queue = Queue.create () in
Queue.add x queue;
while not (Queue.is_empty queue) do
match Queue.take queue with
Empty -> ()
| Node (v,l,r) -> f v;
Queue.add l queue;
Queue.add r queue
done
let tree =
Node (1,
Node (2,
Node (4,
Node (7, Empty, Empty),
Empty),
Node (5, Empty, Empty)),
Node (3,
Node (6,
Node (8, Empty, Empty),
Node (9, Empty, Empty)),
Empty))
let () =
preorder (Printf.printf "%d ") tree; print_newline ();
inorder (Printf.printf "%d ") tree; print_newline ();
postorder (Printf.printf "%d ") tree; print_newline ();
levelorder (Printf.printf "%d ") tree; print_newline ()
Output:
1 2 4 7 5 3 6 8 9 7 4 2 5 1 8 6 9 3 2 4 7 5 3 6 8 9 1 1 2 3 4 5 6 7 8 9
[edit] Oz
declare
Tree = n(1
n(2
n(4 n(7 e e) e)
n(5 e e))
n(3
n(6 n(8 e e) n(9 e e))
e))
fun {Concat Xs}
{FoldR Xs Append nil}
end
fun {Preorder T}
case T of e then nil
[] n(V L R) then
{Concat [[V]
{Preorder L}
{Preorder R}]}
end
end
fun {Inorder T}
case T of e then nil
[] n(V L R) then
{Concat [{Inorder L}
[V]
{Inorder R}]}
end
end
fun {Postorder T}
case T of e then nil
[] n(V L R) then
{Concat [{Postorder L}
{Postorder R}
[V]]}
end
end
local
fun {Collect Queue}
case Queue of nil then nil
[] e|Xr then {Collect Xr}
[] n(V L R)|Xr then
V|{Collect {Append Xr [L R]}}
end
end
in
fun {Levelorder T}
{Collect [T]}
end
end
in
{Show {Preorder Tree}}
{Show {Inorder Tree}}
{Show {Postorder Tree}}
{Show {Levelorder Tree}}
[edit] Python
from collections import namedtuple
from sys import stdout
Node = namedtuple('Node', 'data, left, right')
tree = Node(1,
Node(2,
Node(4,
Node(7, None, None),
None),
Node(5, None, None)),
Node(3,
Node(6,
Node(8, None, None),
Node(9, None, None)),
None))
def printwithspace(i):
stdout.write("%i " % i)
def preorder(node, visitor = printwithspace):
if node is not None:
visitor(node.data)
preorder(node.left, visitor)
preorder(node.right, visitor)
def inorder(node, visitor = printwithspace):
if node is not None:
inorder(node.left, visitor)
visitor(node.data)
inorder(node.right, visitor)
def postorder(node, visitor = printwithspace):
if node is not None:
postorder(node.left, visitor)
postorder(node.right, visitor)
visitor(node.data)
def levelorder(node, more=None, visitor = printwithspace):
if node is not None:
if more is None:
more = []
more += [node.left, node.right]
visitor(node.data)
if more:
levelorder(more[0], more[1:], visitor)
stdout.write(' preorder: ')
preorder(tree)
stdout.write('\n inorder: ')
inorder(tree)
stdout.write('\n postorder: ')
postorder(tree)
stdout.write('\nlevelorder: ')
levelorder(tree)
stdout.write('\n')
Sample output:
preorder: 1 2 4 7 5 3 6 8 9 inorder: 7 4 2 5 1 8 6 9 3 postorder: 7 4 5 2 8 9 6 3 1 levelorder: 1 2 3 4 5 6 7 8 9
[edit] Ruby
class BinaryTreeNode
def initialize(value, left=nil, right=nil)
@value, @left, @right = value, left, right
end
attr_reader :value, :left, :right
def self.from_array(nested_list)
value, left, right = nested_list
if value
self.new(value, self.from_array(left), self.from_array(right))
end
end
def walk_nodes(order, &block)
order.each do |node|
case node
when :left then left && left.walk_nodes(order, &block)
when :self then yield self
when :right then right && right.walk_nodes(order, &block)
end
end
end
def each_preorder(&b) ; walk_nodes([:self, :left, :right], &b) ; end
def each_inorder(&b) ; walk_nodes([:left, :self, :right], &b) ; end
def each_postorder(&b) ; walk_nodes([:left, :right, :self], &b) ; end
def each_levelorder
queue = [self]
until queue.empty?
node = queue.shift
yield node
queue << node.left if node.left
queue << node.right if node.right
end
end
end
root = BinaryTreeNode.from_array [1, [2, [4, 7], [5]], [3, [6, [8], [9]]]]
%w{each_preorder each_inorder each_postorder each_levelorder}.each {|mthd|
printf "%-11s ", mthd[5..-1] + ':'
root.send(mthd) {|node| print "#{node.value} "}
puts
}
Output:
preorder: 1 2 4 7 5 3 6 8 9 inorder: 7 4 2 5 1 8 6 9 3 postorder: 7 4 5 2 8 9 6 3 1 levelorder: 1 2 3 4 5 6 7 8 9
[edit] Tcl
Works with: Tcl version 8.6
oo::class create tree {
# Basic tree data structure stuff...
variable val l r
constructor {value {left {}} {right {}}} {
set val $value
set l $left
set r $right
}
method value {} {return $val}
method left {} {return $l}
method right {} {return $r}
destructor {
if {$l ne ""} {$l destroy}
if {$r ne ""} {$r destroy}
}
# Traversal methods
method preorder {varName script {level 0}} {
upvar [incr level] $varName var
set var $val
uplevel $level $script
if {$l ne ""} {$l preorder $varName $script $level}
if {$r ne ""} {$r preorder $varName $script $level}
}
method inorder {varName script {level 0}} {
upvar [incr level] $varName var
if {$l ne ""} {$l inorder $varName $script $level}
set var $val
uplevel $level $script
if {$r ne ""} {$r inorder $varName $script $level}
}
method postorder {varName script {level 0}} {
upvar [incr level] $varName var
if {$l ne ""} {$l postorder $varName $script $level}
if {$r ne ""} {$r postorder $varName $script $level}
set var $val
uplevel $level $script
}
method levelorder {varName script} {
upvar 1 $varName var
set nodes [list [self]]; # A queue of nodes to process
while {[llength $nodes] > 0} {
set nodes [lassign $nodes n]
set var [$n value]
uplevel 1 $script
if {[$n left] ne ""} {lappend nodes [$n left]}
if {[$n right] ne ""} {lappend nodes [$n right]}
}
}
}
Note that in Tcl it is conventional to handle performing something “for each element” by evaluating a script in the caller's scope for each node after setting a caller-nominated variable to the value for that iteration. Doing this transparently while recursing requires the use of a varying ‘level’ parameter to upvar and uplevel, but makes for compact and clear code.
Demo code to satisfy the official challenge instance:
# Helpers to make construction and listing of a whole tree simpler
proc Tree nested {
lassign $nested v l r
if {$l ne ""} {set l [Tree $l]}
if {$r ne ""} {set r [Tree $r]}
tree new $v $l $r
}
proc Listify {tree order} {
set list {}
$tree $order v {
lappend list $v
}
return $list
}
# Make a tree, print it a few ways, and destroy the tree
set t [Tree {1 {2 {4 7} 5} {3 {6 8 9}}}]
puts "preorder: [Listify $t preorder]"
puts "inorder: [Listify $t inorder]"
puts "postorder: [Listify $t postorder]"
puts "level-order: [Listify $t levelorder]"
$t destroy
Output:
preorder: 1 2 4 7 5 3 6 8 9 inorder: 7 4 2 5 1 8 6 9 3 postorder: 7 4 5 2 8 9 6 3 1 level-order: 1 2 3 4 5 6 7 8 9
[edit] UNIX Shell
Bash (also "sh" on most Unix systems) has arrays. We implement a node as an association between three arrays: left, right, and value.
left=()
right=()
value=()
# node node#, left#, right#, value
#
# if value is empty, use node#
node() {
nx=${1:-'Missing node index'}
leftx=${2}
rightx=${3}
val=${4:-$1}
value[$nx]="$val"
left[$nx]="$leftx"
right[$nx]="$rightx"
}
# define the tree
node 1 2 3
node 2 4 5
node 3 6
node 4 7
node 5
node 6 8 9
node 7
node 8
node 9
# walk NODE# ORDER
walk() {
local nx=${1-"Missing index"}
shift
for branch in "$@" ; do
case "$branch" in
left) if [[ "${left[$nx]}" ]]; then walk ${left[$nx]} $@ ; fi ;;
right) if [[ "${right[$nx]}" ]]; then walk ${right[$nx]} $@ ; fi ;;
self) printf "%d " "${value[$nx]}" ;;
esac
done
}
apush() {
local var="$1"
eval "$var=( \"\${$var[@]}\" \"$2\" )"
}
showname() {
printf "%-12s " "$1:"
}
showdata() {
showname "$1"
shift
walk "$@"
echo ''
}
preorder() { showdata $FUNCNAME $1 self left right ; }
inorder() { showdata $FUNCNAME $1 left self right ; }
postorder() { showdata $FUNCNAME $1 left right self ; }
levelorder() {
showname 'level-order'
queue=( $1 )
x=0
while [[ $x < ${#queue[*]} ]]; do
value="${queue[$x]}"
printf "%d " "$value"
for more in "${left[$value]}" "${right[$value]}" ; do
if [[ -n "$more" ]]; then
apush queue "$more"
fi
done
: $((x++))
done
echo ''
}
preorder 1
inorder 1
postorder 1
levelorder 1
The output:
preorder: 1 2 4 7 5 3 6 8 9
inorder: 7 4 2 5 1 8 6 9 3
postorder: 7 4 5 2 8 9 6 3 1
level-order: 1 2 3 4 5 6 7 8 9
[edit] Ursala
Ursala has built-in notation for trees and is perfect for whipping up little tree walking functions. This source listing shows the tree depicted above declared as a constant, followed by declarations of four functions applicable to trees of any type. The main program applies all four of them to the tree and makes a list of their results, each of which is a list of natural numbers. The compiler directive #cast %nLL induces the compile-time side effect of displaying the result on standard output as a list of lists of naturals.
tree =
1^:<
2^: <4^: <7^: <>, 0>, 5^: <>>,
3^: <6^: <8^: <>, 9^: <>>, 0>>
pre = ~&dvLPCo
post = ~&vLPdNCTo
in = ~&vvhPdvtL2CTiQo
lev = ~&iNCaadSPfavSLiF3RTaq
#cast %nLL
main = <.pre,in,post,lev> tree
output:
< <1,2,4,7,5,3,6,8,9>, <7,4,2,5,1,8,6,9,3>, <7,4,5,2,8,9,6,3,1>, <1,2,3,4,5,6,7,8,9>>
Categories: Programming Tasks | Data Structures | Recursion | Ada | C | C++ | Boost | Clojure | Common Lisp | D | E | Erlang | Factor | Forth | Haskell | J | Java | JavaScript | Logo | OCaml | Oz | Python | Ruby | Tcl | UNIX Shell | Ursala
