# Sorting algorithms/Tree sort on a linked list

Sorting algorithms/Tree sort on a linked list is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Sorting Algorithm
This is a sorting algorithm.   It may be applied to a set of data in order to sort it.     For comparing various sorts, see compare sorts.   For other sorting algorithms,   see sorting algorithms,   or:

O(n logn) sorts

O(n log2n) sorts
Shell Sort

 This page uses content from Wikipedia. The current wikipedia article is at Tree_sort. The original RosettaCode article was extracted from the wikipedia article № 295989333 of 15:13, 12 June 2009 . The list of authors can be seen in the page history. As with Rosetta Code, the pre 5 June 2009 text of Wikipedia is available under the GNU FDL. (See links for details on variance)

A tree sort is a sort algorithm that builds a binary search tree from the keys to be sorted, and then traverses the tree (in-order) so that the keys come out in sorted order. Its typical use is when sorting the elements of a stream from a file. Several other sorts would have to load the elements to a temporary data structure, whereas in a tree sort the act of loading the input into a data structure is sorting it.

The tree sort is considered by some to be the faster method to sort a linked list, followed by Quicksort and Mergesort:

Sediment sort, bubble sort, selection sort perform very badly.

First, construct a doubly linked list (unsorted).
Then construct a tree in situ: use the prev and next of that list as left and right tree pointers.
Then traverse the tree, in order, and recreate a doubly linked list, again in situ, but of course now in sorted order.

## ATS

```(*

Tree sort based on the algorithm at http://archive.today/WM83M

One change is that, instead of a comparison function returning an
integer, we have a template function that serves as order
predicate. In other words, it is a "less than" function.

The mutable structures are implemented in C. The doubly-linked list
implementation is "unsafe". (A "safe" implementation of
doubly-linked lists would be nontrivial.) The ATS code is in an
"imperative" style.

*)

%{^

#include <stddef.h>
#include <stdlib.h>
#include <string.h>

atstype_ptr data;
};

atstype_int count;
};

{
node->data = data;
node->prev = NULL;
node->next = NULL;
return node;
}

static inline atstype_ptr
{
return node->data;
}

{
return node->prev;
}

{
return node->next;
}

static inline void
{
node->prev = new_prev;
}

static inline void
{
node->next = new_next;
}

{
list->tail = NULL;
list->count = 0;
return list;
}

{
}

{
return list->tail;
}

static inline int
{
return list->count;
}

static inline void
{
}

static inline void
{
list->tail = new_tail;
}

static inline void
int new_count)
{
list->count = new_count;
}

%}

abstype dlnode (a : t@ype+, is_nil : bool) = ptr
typedef dlnode (a : t@ype+) = [is_nil : bool] dlnode (a, is_nil)

abstype dllist (a : t@ype+, n : int) = ptr
typedef dllist (a : t@ype+) = [n : int] dllist (a, n)

fn {a : t@ype}
dlnode_make (elem : a) : dlnode (a, false) =
let
extern fn dlnode_make__ : ptr -> ptr = "mac#%"
val data = \$extfcall (ptr, "ATS_MALLOC", sizeof<a>)
val () = \$UN.ptr0_set<a> (data, elem)
in
\$UN.cast (dlnode_make__ data)
end

fn {a : t@ype}
dlnode_nil () : dlnode (a, true) =
\$UN.cast the_null_ptr

fn {}
dlnode_is_nil
{is_nil : bool}
{a      : t@ype}
(node   : dlnode (a, is_nil))
: [b : bool | b == is_nil] bool b =
\$UN.cast (iseqz (\$UN.cast{ptr} node))

fn {}
dlnode_isnot_nil
{is_nil : bool}
{a      : t@ype}
(node   : dlnode (a, is_nil))
: [b : bool | b == ~is_nil] bool b =
\$UN.cast (isneqz (\$UN.cast{ptr} node))

fn {a : t@ype}
dlnode_get_elem (node : dlnode (a, false)) : a =
let
extern fn dlnode_get_data__ : ptr -> ptr = "mac#%"
val data = dlnode_get_data__ (\$UN.cast node)
in
\$UN.ptr0_get<a> data
end

fn {a : t@ype}
dlnode_get_prev (node : dlnode (a, false)) : dlnode a =
let
extern fn dlnode_get_prev__ : ptr -> ptr = "mac#%"
in
\$UN.cast (dlnode_get_prev__ (\$UN.cast node))
end

fn {a : t@ype}
dlnode_get_next (node : dlnode (a, false)) : dlnode a =
let
extern fn dlnode_get_next__ : ptr -> ptr = "mac#%"
in
\$UN.cast (dlnode_get_next__ (\$UN.cast node))
end

fn {a : t@ype}
dlnode_set_prev
(node     : dlnode (a, false),
new_prev : dlnode a)
: void =
let
extern fn dlnode_set_prev__ : (ptr, ptr) -> void = "mac#%"
in
dlnode_set_prev__ (\$UN.cast node, \$UN.cast new_prev)
end

fn {a : t@ype}
dlnode_set_next
(node     : dlnode (a, false),
new_next : dlnode a)
: void =
let
extern fn dlnode_set_next__ : (ptr, ptr) -> void = "mac#%"
in
dlnode_set_next__ (\$UN.cast node, \$UN.cast new_next)
end

fn {a : t@ype}
dllist_make () : dllist (a, 0) =
let
extern fn dllist_make__ : () -> ptr = "mac#%"
in
\$UN.cast (dllist_make__ ())
end

fn {a : t@ype}
dllist_get_head (lst : dllist a) : dlnode a =
let
extern fn dllist_get_head__ : ptr -> ptr = "mac#%"
in
end

fn {a : t@ype}
dllist_get_tail (lst : dllist a) : dlnode a =
let
extern fn dllist_get_tail__ : ptr -> ptr = "mac#%"
in
\$UN.cast (dllist_get_tail__ (\$UN.cast lst))
end

fn {}
dllist_get_count
{n   : int}
{a   : t@ype}
(lst : dllist (a, n))
: int n =
let
extern fn dllist_get_count__ : ptr -> int = "mac#%"
in
\$UN.cast (dllist_get_count__ (\$UN.cast lst))
end

fn {a : t@ype}
(lst      : dllist a,
: void =
let
extern fn dllist_set_head__ : (ptr, ptr) -> void = "mac#%"
in
end

fn {a : t@ype}
dllist_set_tail
(lst      : dllist a,
new_tail : dlnode a)
: void =
let
extern fn dllist_set_tail__ : (ptr, ptr) -> void = "mac#%"
in
dllist_set_tail__ (\$UN.cast lst, \$UN.cast new_tail)
end

fn {a : t@ype}
dllist_set_count {n         : int}
(lst       : &dllist a >> dllist (a, n),
new_count : int n)
: void =
let
extern fn dllist_set_count__ : (ptr, int) -> void = "mac#%"
val () = dllist_set_count__ (\$UN.cast lst, \$UN.cast new_count)
prval () = \$UN.castvwtp2void{dllist (a, n)} lst
in
end

fn {}
dllist_is_empty
{n   : int}
{a   : t@ype}
(lst : dllist (a, n))
: [b : bool | b == (n == 0)] bool b =
dllist_get_count lst = 0

fn {}
dllist_isnot_empty
{n   : int}
{a   : t@ype}
(lst : dllist (a, n))
: [b : bool | b == (n != 0)] bool b =
dllist_get_count lst <> 0

fn {a : t@ype}
dllist_insert_at_end
{n        : int}
(lst      : &dllist (a, n) >> dllist (a, n + 1),
new_elem : a)
: void =
let
val node = dlnode_make<a> new_elem
val n = length lst
in
set_count<a> (lst, succ n);
if n = 0 then
begin
set_tail<a> (lst, node)
end
else
let
val last_node = get_tail<a> lst
val () = assertloc (isneqz last_node)
in
set_next<a> (last_node, node);
set_prev<a> (node, last_node);
set_tail<a> (lst, node)
end
end

infix +=

fn {a : t@ype}
dllist2list {n   : nat}
(lst : dllist (a, n))
: list (a, n) =
let
val n = length lst
fun
loop {i : nat | i <= n}
.<n - i>.
(last_node : dlnode a,
accum     : list (a, i),
i         : int i)
: list (a, n) =
if i = n then
let
val () = assertloc (iseqz last_node)
in
accum
end
else
let
val () = assertloc (isneqz last_node)
val elem = get_elem<a> last_node
in
loop (get_prev<a> last_node,
list_cons (elem, accum),
succ i)
end
in
loop (get_tail<a> lst, list_nil (), 0)
end

fn {a : t@ype}
list2dllist {n   : nat}
(lst : list (a, n))
: dllist (a, n) =
let
fun
loop {i     : nat | i <= n}
.<n - i>.
(lst   : list (a, n - i),
accum : &dllist (a, i) >> dllist (a, n))
: void =
case+ lst of
| list_nil () => ()
| list_cons (elem, rest) =>
begin
accum += elem;
loop (rest, accum)
end

var retval = dllist_make<a> ()
in
loop {0} (lst, retval);
retval
end

extern fn {a : t@ype}           (* The "less than" template. *)
dllist_tree_sort\$lt : (a, a) -> bool

fn {a : t@ype}
dllist2tree {n    : nat}
(lst  : &dllist (a, n) >> _,
root : &dlnode a? >> dlnode a)
: void =
begin
if isneqz root then
let
var node : dlnode a = get_next<a> root
in
set_prev<a> (root, dlnode_nil ());
set_next<a> (root, dlnode_nil ());
while (isneqz node)
let
val next = get_next<a> node
var current : dlnode a = root
var previous : dlnode a = dlnode_nil ()
var node_lt_curr : bool = false
in
while (isneqz current)
begin
previous := current;
node_lt_curr :=
dllist_tree_sort\$lt<a> (get_elem<a> node,
get_elem<a> current);
if node_lt_curr then
current := get_prev<a> current
else
current := get_next<a> current
end;
let
prval () =
\$UN.castvwtp2void{[b : bool] dlnode (a, b)} previous
val () = assertloc (isneqz previous)
in
if node_lt_curr then
set_prev<a> (previous, node)
else
set_next<a> (previous, node)
end;
set_prev<a> (node, dlnode_nil ());
set_next<a> (node, dlnode_nil ());
node := next
end
end
end

fn {a : t@ype}
tree2dllist {n    : nat}
(lst  : &dllist (a, n) >> _,
root : dlnode a)
: void =
let
fun
recurs (lst      : &dllist (a, n) >> _,
root     : dlnode a,
previous : &dlnode a >> _,
count    : &int >> _)
: void =
if isneqz root then
let
val left = get_prev<a> root
and right = get_next<a> root
in
recurs (lst, left, previous, count);
if iseqz (get_prev<a> root) * iseqz (get_head<a> lst) then
begin               (* We are at the first element. *)
set_prev<a> (root, dlnode_nil ())
end
else
let
val () = assertloc (isneqz previous)
in
set_next<a> (previous, root);
set_prev<a> (root, previous)
end;
if succ count = length lst then
begin               (* We are at the last element. *)
set_tail<a> (lst, root);
set_next<a> (root, dlnode_nil ())
end;
previous := root;
count := succ count;
recurs (lst, right, previous, count)
end

var previous : dlnode a = dlnode_nil ()
var count : int = 0
in
set_tail<a> (lst, dlnode_nil ());
recurs (lst, root, previous, count)
end

fn {a : t@ype}
dllist_tree_sort
{n   : nat}
(lst : &dllist (a, n) >> _)
: void =
let
var root : dlnode a
in
dllist2tree (lst, root);
tree2dllist (lst, root)
end

implement
dllist_tree_sort\$lt<int> (x, y) =
x < y

implement
main0 () =
let
var i : int
var data : List0 int = list_nil ()
in
for (i := 1; i <= 20; i := succ i)
data := list_cons (\$extfcall (int, "rand") % 20, data);
let
var lst = list2dllist<int> data
in
println! (dllist2list<int> lst);
dllist_tree_sort<int> lst;
println! (dllist2list<int> lst)
end
end```
Output:
```\$ patscc -DATS_MEMALLOC_GCBDW -O3 tree_sort_task.dats -lgc && ./a.out
16, 12, 6, 0, 6, 3, 19, 10, 7, 2, 1, 9, 12, 6, 15, 13, 15, 17, 6, 3
0, 1, 2, 3, 3, 6, 6, 6, 6, 7, 9, 10, 12, 12, 13, 15, 15, 16, 17, 19```

I see the task used to have something to do with Finnegan's Wake, and with counting cycles, etc. Here I simply sort a list of integers.

It is unlikely, in ATS, that someone would use doubly-linked lists as their canonical linked list implementation. Sorting a singly-linked list this way would be interesting, but I cannot think of a way to do it without allocating new nodes. Of course, a quicksort or mergesort can be done on a singly-linked list without allocating new nodes.

(Obviously, if one is sorting a non-linear linked list, it is in general necessary to allocate new nodes. However, it is not necessary to allocate any temporary nodes.)

## C

```#include <stdio.h>
#include <stdlib.h>
#include <time.h>

void fatal(const char* message) {
fprintf(stderr, "%s\n", message);
exit(1);
}

void* xmalloc(size_t n) {
void* ptr = malloc(n);
if (ptr == NULL)
fatal("Out of memory");
return ptr;
}

typedef struct node_tag {
int item;
struct node_tag* prev;
struct node_tag* next;
} node_t;

void list_initialize(node_t* list) {
list->prev = list;
list->next = list;
}

void list_destroy(node_t* list) {
node_t* n = list->next;
while (n != list) {
node_t* tmp = n->next;
free(n);
n = tmp;
}
}

void list_append_node(node_t* list, node_t* node) {
node_t* prev = list->prev;
prev->next = node;
list->prev = node;
node->prev = prev;
node->next = list;
}

void list_append_item(node_t* list, int item) {
node_t* node = xmalloc(sizeof(node_t));
node->item = item;
list_append_node(list, node);
}

void list_print(node_t* list) {
printf("[");
node_t* n = list->next;
if (n != list) {
printf("%d", n->item);
n = n->next;
}
for (; n != list; n = n->next)
printf(", %d", n->item);
printf("]\n");
}

void tree_insert(node_t** p, node_t* n) {
while (*p != NULL) {
if (n->item < (*p)->item)
p = &(*p)->prev;
else
p = &(*p)->next;
}
*p = n;
}

void tree_to_list(node_t* list, node_t* node) {
if (node == NULL)
return;
node_t* prev = node->prev;
node_t* next = node->next;
tree_to_list(list, prev);
list_append_node(list, node);
tree_to_list(list, next);
}

void tree_sort(node_t* list) {
node_t* n = list->next;
if (n == list)
return;
node_t* root = NULL;
while (n != list) {
node_t* next = n->next;
n->next = n->prev = NULL;
tree_insert(&root, n);
n = next;
}
list_initialize(list);
tree_to_list(list, root);
}

int main() {
srand(time(0));
node_t list;
list_initialize(&list);
for (int i = 0; i < 16; ++i)
list_append_item(&list, rand() % 100);
printf("before sort: ");
list_print(&list);
tree_sort(&list);
printf(" after sort: ");
list_print(&list);
list_destroy(&list);
return 0;
}
```
Output:
```before sort: [33, 57, 20, 49, 32, 48, 13, 81, 18, 76, 98, 47, 11, 4, 21, 5]
after sort: [4, 5, 11, 13, 18, 20, 21, 32, 33, 47, 48, 49, 57, 76, 81, 98]
```

## FreeBASIC

Translation of: Yabasic
```#define key   0
#define izda  1
#define dcha  2

Dim Shared As Integer index, size
index = 0 : size = 10

Dim Shared As String tree(size)
Dim Shared As Integer indTree(size, 3)

Declare Sub insertNode(word As String, prev As Integer)
Declare Sub makeNode(prev As Integer, branch As Integer, word As String)

Function Token(Texto As String, Delim As String, Direcc As Byte = 0) As String
Dim As Integer LocA = Instr(Texto, Delim)
Return Iif(Direcc <= 0, Left(Texto, LocA), Right(Texto, Len(Texto) - LocA))
End Function

Sub makeNode(prev As Integer, branch As Integer, word As String)
If indTree(prev, branch) = 0 Then
index += 1
If index > size Then size += 10 : Redim tree(size) : Redim indTree(size, 3)
indTree(prev, branch) = index
tree(index) = word
indTree(index, key) = 1
Else
insertNode(word, indTree(prev, branch))
End If
End Sub

Sub insertNode(word As String, prev As Integer)
Dim As String pal, ant
pal = Lcase(word)
ant = Lcase(tree(prev))

If ant <> "" Then
If pal < ant Then
makeNode(prev, izda, word)
Elseif pal > ant Then
makeNode(prev, dcha, word)
Elseif pal = ant Then
indTree(prev, key) += 1
End If
Else
index += 1
tree(index) = word
indTree(index, key) = 1
End If
End Sub

Sub showTree(numreg As Integer)
If indTree(numreg, izda) Then showTree(indTree(numreg, izda))
Print tree(numreg); " ";
If indTree(numreg, dcha) Then showTree(indTree(numreg, dcha))
End Sub

Sub makeTree(texto() As String)
For n As Integer = 1 To Ubound(texto)
insertNode(texto(n), 1)
Next n
End Sub

Dim As String g(1 To 10) = {"one","two","three","four","five","six","seven","eight","nine","ten"}
makeTree(g())
showTree(1)
Print
Sleep
```
Output:
`eight five four nine one seven six ten three two`

## Go

This is based on the Kotlin entry but has been adjusted to satisfy the revised task description.

```package main

import (
"container/list"
"fmt"
)

type BinaryTree struct {
node         int
leftSubTree  *BinaryTree
rightSubTree *BinaryTree
}

func (bt *BinaryTree) insert(item int) {
if bt.node == 0 {
bt.node = item
bt.leftSubTree = &BinaryTree{}
bt.rightSubTree = &BinaryTree{}
} else if item < bt.node {
bt.leftSubTree.insert(item)
} else {
bt.rightSubTree.insert(item)
}
}

func (bt *BinaryTree) inOrder(ll *list.List) {
if bt.node == 0 {
return
}
bt.leftSubTree.inOrder(ll)
ll.PushBack(bt.node)
bt.rightSubTree.inOrder(ll)
}
func treeSort(ll *list.List) *list.List {
searchTree := &BinaryTree{}
for e := ll.Front(); e != nil; e = e.Next() {
i := e.Value.(int)
searchTree.insert(i)
}
ll2 := list.New()
searchTree.inOrder(ll2)
return ll2
}

func printLinkedList(ll *list.List, f string, sorted bool) {
for e := ll.Front(); e != nil; e = e.Next() {
i := e.Value.(int)
fmt.Printf(f+" ", i)
}
if !sorted {
fmt.Print("-> ")
} else {
fmt.Println()
}
}

func main() {
sl := []int{5, 3, 7, 9, 1}
ll := list.New()
for _, i := range sl {
ll.PushBack(i)
}
lls := treeSort(ll)

sl2 := []int{'d', 'c', 'e', 'b', 'a'}
ll2 := list.New()
for _, c := range sl2 {
ll2.PushBack(c)
}
lls2 := treeSort(ll2)
}
```
Output:
```5 3 7 9 1 -> 1 3 5 7 9
d c e b a -> a b c d e
```

Due to pure functional nature of Haskell sorting in situ is impossible.

Here we use abstractions of ``Foldable`` type class in order to traverse both the doubly-linked list and the binary tree. Implementation of doubly-linked list is given here Doubly-linked_list/Traversal#Haskell

```{-# language DeriveFoldable #-}
import Data.Foldable

data DList a = End | Elem { prev :: DList a
, elt :: a
, next :: DList a }

mkDList :: Foldable t => t a -> DList a
mkDList = go End . toList
where go _    []     = End
go prev (x:xs) = current
where current = Elem prev x next
next    = go current xs

instance Foldable DList where
foldMap f End = mempty
foldMap f dl = f (elt dl) <> foldMap f (next dl)

sortDL :: Ord a => DList a -> DList a
sortDL = mkDList . mkTree

-- binary tree
data BTree a = Empty | Node { left  :: BTree a
, node  :: a
, right :: BTree a }
deriving (Show, Foldable)

addTree Empty x = Node Empty x Empty
addTree (Node l a g) x =
case compare x a of
LT -> Node (addTree l x) a g
_  -> Node l a (addTree g x)

mkTree :: (Foldable t, Ord a) => t a -> BTree a

treeSort :: (Foldable t, Ord a) => t a -> [a]
treeSort = toList . mkTree
```
```λ> let l = mkDList [2,4,3,5,7,6,2,9]
λ> l
mkDList [2,4,3,5,7,6,2,9] :: Num a => DList a

λ> toList l
[2,4,3,5,7,6,2,9]

λ> mkTree l
Node {left = Empty, node = 2, right = Node {left = Node {left = Node {left = Empty, node = 2, right = Empty}, node = 3, right = Empty}, node = 4, right = Node {left = Empty, node = 5, right = Node {left = Node {left = Empty, node = 6, right = Empty}, node = 7, right = Node {left = Empty, node = 9, right = Empty}}}}}

λ> toList \$ mkTree l
[2,2,3,4,5,6,7,9]

λ> toList \$ sortDL l
[2,2,3,4,5,6,7,9]

λ> treeSort [2,4,3,5,7,6,2,9]
[2,2,3,4,5,6,7,9]```

## J

What *is* a sentence in Finnegan's Wake? Let's say that it's all the text leading up to a period, question mark or exclamation point if (and only if) the character is followed by a space or newline. (There are some practical difficulties here - this means, for example, that the first sentence of a chapter includes the chapter heading - but it's good enough for now.)

There's also the issue of how do we want to sort the sentences? Let's say we'll sort them in ascii order without normalization of the text (since that is simplest).

Let's also say that we have prepared a file which contains some sort of ascii rendition of the text. Note that the final result we get here will depend on exactly how that ascii rendition was prepared. But let's just ignore that issue so we can get something working.

Next, we need to think of what kind of tree, there are a great number of kinds of trees, and they can be categorized in many different ways. For example, a directory tree is almost never a balanced binary tree. (Note that a linked list is a kind of a tree - an extremely tall and skinny unbalanced tree, but a tree nonetheless - and a binary tree at that. Then again, note that efficiency claims in general are specious, because efficient for one purpose tends to be inefficient for many other purposes.) Since we are going for efficiency here, we will implement a short, fat tree (let's call that "efficient use of the programmer's time" or something like that...). Specifically, we'll be implementing a one level deep tree which happens to have 14961 leaves connected directly to the root node. (Edit: task description has been changed to mandate a specific binary tree. But we are going to ignore that here, since the consequence would be several orders of magnitude slowdown, and a lot of extra code to write. That kind of detail can be useful in an educational setting, and in some technology settings, but it would cause real problems here.)

Simplicity is a virtue, right?

Finally, there's the matter of counting swaps. Let's define our swap count as the minimal number of swaps which would be needed to produce our sorted result.

With these choices, the task becomes:

```   finn=: fread '~user/temp/wake/finneganswake.txt'
sentences=: (<;.2~ '. '&E.@rplc&(LF,' !.?.')) finn
#sentences
14961
+/<:#@>C./:sentences
14945
```

We have to swap almost every sentence, but 16 of them can be sorted "for free" with the swaps of the other sentences.

For that matter, inspecting the lengths of the cycles formed by the minimal arrangements of swaps...

```   /:~ #@>C./:sentences
1 1 2 2 4 9 12 25 32 154 177 570 846 935 1314 10877
```

... we can see that two of the sentences were fine right where they start out. Let's see what they are:

```   ;:inv (#~(= /:~))sentences
Very, all
fourlike tellt.  What tyronte power!
```

So now you know.

(Processing time here is negligible - other than the time needed to fetch a copy of the book and render it as plain text ascii - but if we were careful to implement the efficiency recommendations of this task more in the spirit of whatever the task is presumably implying, we could probably increase the processing time by several orders of magnitude.)

So... ok... let's do this "right" (which is to say according to the current task specification, as opposed to the task specification that was present for the early drafts - though, perhaps, using Finnegan's Wake as a data set encourages a certain degree of ... informality?).

Anyways, here we go:

```left=: i.0
right=: i.0
data=: i.0

insert=:3 :0"0
k=. 0
assert. (left =&# right) * (left =&# data)
if. 0<#data do.
while. k<#data do.
if. y=k{data do.return.end.
n=. k
if. y<k{data do.
k=. k{".p=.'left'
else.
k=. k{".p=.'right'
end.
end.
(p)=:(#data) n} ".p
end.
left=:left, _
right=:right, _
data=:data,y
i.0 0
)

flatten=:3 :0
extract 0
)

extract=:3 :0
if. y>:#data do.'' return. end.
(extract y{left),(y{data),extract y{right
)
```

Example use would be something like:

```   insert sentences
extract''
```

But task's the current url for Finnegan's Wake does not point at flat text and constructing such a thing would be a different task...

## Java

```// TreeSortTest.java
import java.util.*;

public class TreeSortTest {
public static void main(String[] args) {
test1();
System.out.println();
test2();
}

// Sort a random list of integers
private static void test1() {
Random r = new Random();
for (int i = 0; i < 16; ++i)
System.out.println("before sort: " + list);
list.treeSort();
System.out.println(" after sort: " + list);
}

// Sort a list of strings
private static void test2() {
String[] strings = { "one", "two", "three", "four", "five",
"six", "seven", "eight", "nine", "ten"};
for (String str : strings)
System.out.println("before sort: " + list);
list.treeSort();
System.out.println(" after sort: " + list);
}
}
```
```// LinkedList.java

// Java provides a doubly-linked list implementation but it doesn't permit
// public access to its internal structure for obvious reasons, so to
// fulfil the task requirements we must implement one ourselves.

import java.util.*;

public class LinkedList<T extends Comparable<? super T>> {
private final Node<T> sentinel = new Node<T>(null);

clear();
}

public void clear() {
sentinel.next = sentinel;
sentinel.prev = sentinel;
}

public boolean isEmpty() {
return sentinel.next == sentinel;
}

}

n.prev = sentinel.prev;
n.next = sentinel;
sentinel.prev.next = n;
sentinel.prev = n;
}

public String toString() {
StringBuilder str = new StringBuilder("[");
Node<T> n = sentinel.next;
if (n != sentinel) {
str.append(n.item);
n = n.next;
while (n != sentinel) {
str.append(", ");
str.append(n.item);
n = n.next;
}
}
str.append("]");
return str.toString();
}

public void treeSort() {
if (isEmpty())
return;
Node<T> n = sentinel.next;
Node<T> root = null;
while (n != sentinel) {
Node<T> next = n.next;
n.next = null;
n.prev = null;
root = treeInsert(root, n);
n = next;
}
clear();
treeToList(root);
}

private Node<T> treeInsert(Node<T> tree, Node<T> n) {
if (tree == null)
tree = n;
else if (n.item.compareTo(tree.item) < 0)
tree.prev = treeInsert(tree.prev, n);
else
tree.next = treeInsert(tree.next, n);
return tree;
}

private void treeToList(Node<T> node) {
if (node == null)
return;
Node<T> prev = node.prev;
Node<T> next = node.next;
treeToList(prev);
treeToList(next);
}

private static class Node<T> {
private T item;
private Node<T> prev = null;
private Node<T> next = null;

private Node(T item) {
this.item = item;
}
}
}
```
Output:
```before sort: [37, 88, 13, 18, 72, 77, 29, 93, 21, 97, 37, 42, 67, 22, 29, 2]
after sort: [2, 13, 18, 21, 22, 29, 29, 37, 37, 42, 67, 72, 77, 88, 93, 97]

before sort: [one, two, three, four, five, six, seven, eight, nine, ten]
after sort: [eight, five, four, nine, one, seven, six, ten, three, two]
```

## Julia

```mutable struct BTree{T}
data::T
left::Union{BTree, Nothing}
right::Union{BTree, Nothing}
BTree(val::T) where T = new{T}(val, nothing, nothing)
end

function insert(tree, data)
if data < tree.data
if tree.left == nothing
tree.left = BTree(data)
else
insert(tree.left, data)
end
else
if tree.right == nothing
tree.right = BTree(data)
else
insert(tree.right, data)
end
end
end

function sorted(tree)
return tree == nothing ? [] :
typeof(tree.data)[sorted(tree.left); tree.data; sorted(tree.right)]
end

function arraytotree(arr)
tree = BTree(arr[1])
for data in arr[2:end]
insert(tree, data)
end
return tree
end

function testtreesort(arr)
println("Unsorted: ", arr)
tree = arraytotree(arr)
println("Sorted: ", sorted(tree))
end

testtreesort(rand(1:99, 12))
```
Output:
```Unsorted: [1, 12, 15, 22, 28, 26, 69, 22, 1, 62, 73, 95]
Sorted: [1, 1, 12, 15, 22, 22, 26, 28, 62, 69, 73, 95]
```

## Kotlin

As I can't be bothered to download Finnegan's Wake and deal with the ensuing uncertainties, I've contented myself by following a similar approach to the Racket and Scheme entries:

```// version 1.1.51

class BinaryTree<T : Comparable<T>> {
var node: T? = null
lateinit var leftSubTree: BinaryTree<T>
lateinit var rightSubTree: BinaryTree<T>

fun insert(item: T) {
if (node == null) {
node = item
leftSubTree = BinaryTree<T>()
rightSubTree = BinaryTree<T>()
}
else if (item < node as T) {
leftSubTree.insert(item)
}
else {
rightSubTree.insert(item)
}
}

fun inOrder() {
if (node == null) return
leftSubTree.inOrder()
print("\$node ")
rightSubTree.inOrder()
}
}

fun <T : Comparable<T>> LinkedList<T>.treeSort() {
val searchTree = BinaryTree<T>()
for (item in this) searchTree.insert(item)
print("\${this.joinToString(" ")} -> ")
searchTree.inOrder()
println()
}

fun main(args: Array<String>) {
val ll = LinkedList(listOf(5, 3, 7, 9, 1))
ll.treeSort()
val ll2 = LinkedList(listOf('d', 'c', 'e', 'b' , 'a'))
ll2.treeSort()
}
```
Output:
```5 3 7 9 1 -> 1 3 5 7 9
d c e b a -> a b c d e
```

## Nim

Inspired by C and Java solutions.

As Nim standard library provides a doubly linked list implementation which allows to access to the "prev" and "next" fields, we used it. So, we had only to write the transformation from list to tree and conversely.

```import lists, random

if tree.isNil: tree = node
elif node.value < tree.value: tree.prev.treeInsert(node)
else: tree.next.treeInsert(node)

if node.isNil: return
let prev = node.prev
let next = node.next
list.listFromTree(prev)
list.append(node)
list.listFromTree(next)

var list = list
if list.head == list.tail: return list
while not n.isNil:
var next = n.next
n.next = nil
n.prev = nil
root.treeInsert(n)
n = next
result.listFromTree(root)

randomize()
for i in 0..15: list1.append(rand(10..99))
echo "Before sort: ", list1
echo "After sort:  ", list1.treeSort()
echo()

for s in ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"]:
list2.append(s)
echo "Before sort: ", list2
echo "After sort:  ", list2.treeSort()
```
Output:
```Before sort: [27, 33, 51, 66, 34, 56, 81, 78, 32, 63, 78, 48, 71, 66, 30, 49]
After sort:  [27, 30, 32, 33, 34, 48, 49, 51, 56, 63, 66, 66, 71, 78, 78, 81]

Before sort: ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"]
After sort:  ["eight", "five", "four", "nine", "one", "seven", "six", "ten", "three", "two"]```

## Ol

Ol has builtin sorted key-value trees named "ff". We converting list into ff and back again as already sorted list. Only values (small integers, constants) and symbols are allowed.

```(define (tree-sort l)
(map car (ff->list
(fold (lambda (ff p)
(put ff p #t))
#empty l))))

(print (tree-sort '(5 3 7 9 1)))
```
Output:
```(1 3 5 7 9)
```

## Phix

### version 1

Translation of: Kotlin
```with javascript_semantics

enum KEY,LEFT,RIGHT
function tree_insert(object node, item)
if node=NULL then
node = {item,NULL,NULL}
else
node = deep_copy(node,1) -- (one level only needed)
if item<node[KEY] then
node[LEFT] = tree_insert(node[LEFT],item)
else
node[RIGHT] = tree_insert(node[RIGHT],item)
end if
end if
return node
end function

function inOrder(object node)
sequence res = ""
if node!=NULL then
res = inOrder(node[LEFT])
res &= node[KEY]
res &= inOrder(node[RIGHT])
end if
return res
end function

procedure treeSort(sequence s)
object tree = NULL
for i=1 to length(s) do tree = tree_insert(tree,s[i]) end for
pp({s," => ",inOrder(tree)})
end procedure

treeSort({5, 3, 7, 9, 1})
treeSort("dceba")
```
Output:
```{{5,3,7,9,1}, " => ", {1,3,5,7,9}}
{"dceba", " => ", "abcde"}
```

### version 2

Following my idea of a revised task description, see talk page.

```with javascript_semantics

enum NEXT,PREV,DATA
constant empty_dll = {{1,1}}
sequence dll

procedure insert_after(object data, integer pos=1)
integer prv = dll[pos][PREV]
dll = append(dll,{pos,prv,data})
if prv!=0 then
dll[prv][NEXT] = length(dll)
end if
dll[pos][PREV] = length(dll)
end procedure

procedure append_node(integer node)
-- (like insert_after, but in situ rebuild)
integer prev = dll[1][PREV]
dll[node][NEXT] = 1
dll[node][PREV] = prev
dll[prev][NEXT] = node
dll[1][PREV] = node
end procedure

function dll_collect()
sequence res = ""
integer idx = dll[1][NEXT]
while idx!=1 do
res = append(res,dll[idx][DATA])
idx = dll[idx][NEXT]
end while
return res
end function

-- tree:
enum LEFT,RIGHT,KEY

function tree_insert(integer root, object item, integer idx)
if root=NULL then
return idx
else
integer branch = iff(item<dll[root][KEY]?LEFT:RIGHT)
dll[root][branch] = tree_insert(dll[root][branch],item,idx)
return root
end if
end function

procedure traverse(integer node)
if node!=NULL then
traverse(dll[node][LEFT])
integer right = dll[node][RIGHT]
append_node(node)
traverse(right)
end if
end procedure

bool detailed = true
procedure treeSort()
if detailed then
?{"initial dll",dll}
end if
integer tree = NULL,
idx = dll[1][NEXT]
while idx!=1 do
integer next = dll[idx][NEXT]
dll[idx][NEXT] = NULL
dll[idx][PREV] = NULL
tree = tree_insert(tree,dll[idx][DATA],idx)
idx = next
end while
dll[1] = {tree,0} -- (0 is meaningless, but aligns output)
if detailed then
?{"tree insitu",dll}
end if
dll[1] = deep_copy(empty_dll[1])
traverse(tree)
if detailed then
?{"rebuilt dll",dll}
end if
end procedure

procedure test(sequence s)
dll = deep_copy(empty_dll)
for i=1 to length(s) do insert_after(s[i]) end for
?{"unsorted",dll_collect()}
treeSort()
?{"sorted",dll_collect()}
end procedure

test({5, 3, 7, 9, 1})
detailed = false
test("dceba")
test({"d","c","e","b","a"})
```
Output:
```{"unsorted",{5,3,7,9,1}}
{"initial dll",{{2,6},{3,1,5},{4,2,3},{5,3,7},{6,4,9},{1,5,1}}}
{"tree insitu",{{2,0},{3,4,5},{6,0,3},{0,5,7},{0,0,9},{0,0,1}}}
{"rebuilt dll",{{6,5},{4,3,5},{2,6,3},{5,2,7},{1,4,9},{3,1,1}}}
{"sorted",{1,3,5,7,9}}
{"unsorted","dceba"}
{"sorted","abcde"}
{"unsorted",{"d","c","e","b","a"}}
{"sorted",{"a","b","c","d","e"}}
```

## Racket

Translation of: Scheme

-- this implementation illustrates differences in identifiers and syntaxes of Scheme and Racket's `match-lambda` family. `racket/match` documented here.

```#lang racket/base
(require racket/match)

(define insert
;; (insert key tree)
(match-lambda**
[(x '())         `(() ,x ())]
[(x '(() () ())) `(() ,x ())]
[(x `(,l ,k ,r)) #:when (<= x k) `(,(insert x l) ,k ,r)]
[(x `(,l ,k ,r)) `(,l ,k ,(insert x r))]
[(_ _) "incorrect arguments or broken tree"]))

(define in-order
;; (in-order tree)
(match-lambda
[`(() ,x ()) `(,x)]
[`(,l ,x ())  (append (in-order l) `(,x))]
[`(() ,x ,r)  (append `(,x) (in-order r))]
[`(,l ,x ,r)  (append (in-order l) `(,x) (in-order r))]
[_ "incorrect arguments or broken tree"]))

(define (tree-sort lst)
(define tree-sort-itr
(match-lambda**
[(x `())        (in-order x)]
[(x `(,a . ,b)) (tree-sort-itr (insert a x) b)]
[(_ _) "incorrect arguments or broken tree"]))
(tree-sort-itr '(() () ()) lst))

(tree-sort '(5 3 7 9 1))
```
Output:
`'(1 3 5 7 9)`

## Raku

Translation of: Go
```# 20231201 Raku programming solution

class BinaryTree { has (\$.node, \$.leftSubTree, \$.rightSubTree) is rw;

method insert(\$item) {
if not \$.node.defined {
\$.node = \$item;
(\$.leftSubTree, \$.rightSubTree)>>.&{ \$_ = BinaryTree.new }
} elsif \$item cmp \$.node < 0 {
\$.leftSubTree.insert(\$item);
} else {
\$.rightSubTree.insert(\$item);
}
}

method inOrder(@ll) {
return unless \$.node.defined;
\$.leftSubTree.inOrder(@ll);
@ll.push(\$.node);
\$.rightSubTree.inOrder(@ll);
}
}

sub treeSort(@ll) {
my \$searchTree = BinaryTree.new;
for @ll -> \$i { \$searchTree.insert(\$i) }
\$searchTree.inOrder(my @ll2);
return @ll2
}

sub printLinkedList(@ll, Str \$fmt, Bool \$sorted) {
for @ll -> \$i { printf "\$fmt ", \$i }
\$sorted ?? say() !! print "-> "
}

my @ll  = <5 3 7 9 1>;
#my @ll = [37, 88, 13, 18, 72, 77, 29, 93, 21, 97, 37, 42, 67, 22, 29, 2];
my @lls = treeSort(@ll);

my @ll2 = <d c e b a>;
#my @ll2 = <one two three four five six seven eight nine ten>;
my @lls2 = treeSort(@ll2);
```

You may Attempt This Online!

## Scheme

The following implements a sorting algorithm that takes a linked list, puts each key into an unbalanced binary tree and returns an in-order traversal of the tree.

Library: Matchable
Works with: Chicken Scheme
```(use matchable)

(define insert
;; (insert key tree)
(match-lambda*
[(x ())         `(() ,x ()) ]
[(x (() () ())) `(() ,x ()) ]
[(x (l k r))
(=> continue)
(if (<= x k)
`(,(insert x l) ,k ,r)
(continue)) ]
[(x (l k r)) `(,l ,k ,(insert x r)) ]
[_ "incorrect arguments or broken tree" ]))

(define in-order
;; (in-order tree)
(match-lambda
[(() x ()) `(,x)]
[(l x ())  (append (in-order l) `(,x))]
[(() x r)  (append `(,x) (in-order r))]
[(l x r)   (append (in-order l) `(,x) (in-order r))]
[_ "incorrect arguments or broken tree" ]))

(define (tree-sort lst)
(define tree-sort-itr
(match-lambda*
[(x ())      (in-order x)]
[(x (a . b)) (tree-sort-itr (insert a x) b)]
[_ "incorrect arguments or broken tree" ]))
(tree-sort-itr '(() () ()) lst))
```

Usage:

``` #;2> (tree-sort '(5 3 7 9 1))
(1 3 5 7 9)
```

## Wren

Translation of: Kotlin
Library: Wren-llist
Library: wren-sort
```import "./llist" for DLinkedList
import "./sort" for Cmp

class BinaryTree {
construct new() {
_node = null
_leftSubTree  = null
_rightSubTree = null
}

insert(item) {
if (!_node) {
_node = item
_leftSubTree  = BinaryTree.new()
_rightSubTree = BinaryTree.new()
} else {
var cmp = Cmp.default(item)
if (cmp.call(item, _node) < 0) {
_leftSubTree.insert(item)
} else {
_rightSubTree.insert(item)
}
}
}

inOrder() {
if (!_node) return
_leftSubTree.inOrder()
System.write("%(_node) ")
_rightSubTree.inOrder()
}
}

var treeSort = Fn.new { |ll|
var searchTree = BinaryTree.new()
for (item in ll) searchTree.insert(item)
System.write("%(ll.join(" ")) -> ")
searchTree.inOrder()
System.print()
}

var ll = DLinkedList.new([5, 3, 7, 9, 1])
treeSort.call(ll)
var ll2 = DLinkedList.new(["d", "c", "e", "b", "a"])
treeSort.call(ll2)
```
Output:
```5 3 7 9 1 -> 1 3 5 7 9
d c e b a -> a b c d e
```

## Yabasic

```// Rosetta Code problem: http://rosettacode.org/wiki/Tree_sort_on_a_linked_list
// by Galileo, 04/2022

clear screen

KEY = 0 : LEFT = 1 : RIGHT = 2

index = 0 : size = 10

dim tree\$(size)
dim indTree(size, 3)

sub makeNode(prev, branch, word\$)
if indTree(prev, branch) = 0 then
index = index + 1
if index > size then size = size + 10 : redim tree\$(size) : redim indTree(size, 3) end if
indTree(prev, branch) = index
tree\$(index) = word\$
indTree(index, KEY) = 1
else
insertNode(word\$, indTree(prev, branch))
end if
end sub

sub insertNode(word\$, prev)
local pal\$, ant\$

pal\$ = lower\$(word\$)
ant\$ = lower\$(tree\$(prev))

if ant\$ <> "" then
if pal\$ < ant\$ then
makeNode(prev, LEFT, word\$)
elseif pal\$ > ant\$ then
makeNode(prev, RIGHT, word\$)
elseif pal\$ = ant\$ then
indTree(prev, KEY) = indTree(prev, KEY) + 1
end if
else
index = index + 1
tree\$(index) = word\$
indTree(index,KEY) = 1
end if
end sub

sub showTree(numreg)

if indTree(numreg,LEFT) then
showTree(indTree(numreg, LEFT))
end if
print tree\$(numreg), " ";
if indTree(numreg, RIGHT) then
showTree(indTree(numreg, RIGHT))
end if
end sub

sub makeTree(line\$)
local n, numwords, words\$(1)

numwords = token(line\$, words\$())

for n = 1 to numwords
insertNode(words\$(n), 1)
next n
end sub

makeTree("one two three four five six seven eight nine ten")
showTree(1)
print```
Output:
```eight five four nine one seven six ten three two
---Program done, press RETURN---```

## zkl

This code reads a file [of source code] line by line, and builds a binary tree of the first word of each line. Then prints the sorted list.

```class Node{
var left,right,value;
fcn init(value){ self.value=value; }
}
class Tree{
var root;
if(not root){ root=Node(value); return(self); }
fcn(node,value){
if(not node) return(Node(value));
if(value!=node.value){  // don't add duplicate values
if(value<node.value) node.left =self.fcn(node.left, value);
else                 node.right=self.fcn(node.right,value);
}
node
}(root,value);
return(self);
}
fcn walker{ Utils.Generator(walk,root); }
fcn walk(node){	// in order traversal
if(node){
self.fcn(node.left);
vm.yield(node.value);
self.fcn(node.right);
}
}
}```
```tree:=Tree();
File("bbb.zkl").pump(tree.add,fcn(line){  // 5,000 lines to 660 words
line.split(" ")[0].strip();	// take first word
});

foreach word in (tree){ println(word) }```
Output:
```...
Atomic.sleep(0.5);
Atomic.sleep(100000);
Atomic.sleep(2);
Atomic.waitFor(fcn{
Boyz:=Boys.pump(D(),fcn([(b,gs)]){
Compiler.Compiler.compileText(code)();
...
```