Sorting algorithms/Tree sort on a linked list

From Rosetta Code
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.

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.


Task:
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.

*)

#define ATS_EXTERN_PREFIX "tree_sort_task_"

#include "share/atspre_staload.hats"
staload UN = "prelude/SATS/unsafe.sats"

%{^

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

struct tree_sort_task_dlnode {
  atstype_ptr data;
  struct tree_sort_task_dlnode *prev;
  struct tree_sort_task_dlnode *next;
};

typedef struct tree_sort_task_dlnode *tree_sort_task_dlnode_t;

struct tree_sort_task_dllist {
  tree_sort_task_dlnode_t head;
  tree_sort_task_dlnode_t tail;
  atstype_int count;
};

typedef struct tree_sort_task_dllist *tree_sort_task_dllist_t;

static tree_sort_task_dlnode_t
tree_sort_task_dlnode_make__ (atstype_ptr data)
{
  tree_sort_task_dlnode_t node =
    ATS_MALLOC (sizeof (struct tree_sort_task_dlnode));
  node->data = data;
  node->prev = NULL;
  node->next = NULL;
  return node;
}

static inline atstype_ptr
tree_sort_task_dlnode_get_data__ (tree_sort_task_dlnode_t node)
{
  return node->data;
}

static inline tree_sort_task_dlnode_t
tree_sort_task_dlnode_get_prev__ (tree_sort_task_dlnode_t node)
{
  return node->prev;
}

static inline tree_sort_task_dlnode_t
tree_sort_task_dlnode_get_next__ (tree_sort_task_dlnode_t node)
{
  return node->next;
}

static inline void
tree_sort_task_dlnode_set_prev__ (tree_sort_task_dlnode_t node,
                                  tree_sort_task_dlnode_t new_prev)
{
  node->prev = new_prev;
}

static inline void
tree_sort_task_dlnode_set_next__ (tree_sort_task_dlnode_t node,
                                  tree_sort_task_dlnode_t new_next)
{
  node->next = new_next;
}

static tree_sort_task_dllist_t
tree_sort_task_dllist_make__ (void)
{
  tree_sort_task_dllist_t list =
    ATS_MALLOC (sizeof (struct tree_sort_task_dllist));
  list->head = NULL;
  list->tail = NULL;
  list->count = 0;
  return list;
}

static inline tree_sort_task_dlnode_t
tree_sort_task_dllist_get_head__ (tree_sort_task_dllist_t list)
{
  return list->head;
}

static inline tree_sort_task_dlnode_t
tree_sort_task_dllist_get_tail__ (tree_sort_task_dllist_t list)
{
  return list->tail;
}

static inline int
tree_sort_task_dllist_get_count__ (tree_sort_task_dllist_t list)
{
  return list->count;
}

static inline void
tree_sort_task_dllist_set_head__ (tree_sort_task_dllist_t list,
                                  tree_sort_task_dlnode_t new_head)
{
  list->head = new_head;
}

static inline void
tree_sort_task_dllist_set_tail__ (tree_sort_task_dllist_t list,
                                  tree_sort_task_dlnode_t new_tail)
{
  list->tail = new_tail;
}

static inline void
tree_sort_task_dllist_set_count__ (tree_sort_task_dllist_t list,
                                   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

overload iseqz with dlnode_is_nil
overload isneqz with dlnode_isnot_nil
overload get_elem with dlnode_get_elem
overload get_prev with dlnode_get_prev
overload get_next with dlnode_get_next
overload set_prev with dlnode_set_prev
overload set_next with dlnode_set_next

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
    $UN.cast (dllist_get_head__ ($UN.cast lst))
  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}
dllist_set_head
          (lst      : dllist a,
           new_head : dlnode a)
    : void =
  let
    extern fn dllist_set_head__ : (ptr, ptr) -> void = "mac#%"
  in
    dllist_set_head__ ($UN.cast lst, $UN.cast new_head)
  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

overload length with dllist_get_count
overload iseqz with dllist_is_empty
overload isneqz with dllist_isnot_empty
overload get_head with dllist_get_head
overload get_tail with dllist_get_tail
overload get_count with dllist_get_count
overload set_head with dllist_set_head
overload set_tail with dllist_set_tail
overload set_count with dllist_set_count

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_head<a> (lst, node);
        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 +=
overload += with dllist_insert_at_end

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
    root := get_head lst;
    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_head<a> (lst, root);
              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_head<a> (lst, dlnode_nil ());
    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

Some comments:

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)
    }
    printLinkedList(ll, "%d", false)
    lls := treeSort(ll)
    printLinkedList(lls, "%d", true)

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

Haskell

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

-- double-linked list
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
mkTree = foldl addTree Empty

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
)

This could be wrapped differently, but it's adequate for this task.

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() {
        LinkedList<Integer> list = new LinkedList<>();
        Random r = new Random();
        for (int i = 0; i < 16; ++i)
            list.add(Integer.valueOf(r.nextInt(100)));
        System.out.println("before sort: " + list);
        list.treeSort();
        System.out.println(" after sort: " + list);
    }

    // Sort a list of strings
    private static void test2() {
        LinkedList<String> list = new LinkedList<>();
        String[] strings = { "one", "two", "three", "four", "five",
            "six", "seven", "eight", "nine", "ten"};
        for (String str : strings)
            list.add(str);
        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);

    public LinkedList() {
        clear();
    }

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

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

    public void add(T item) {
        addNode(new Node<T>(item));
    }

    private void addNode(Node<T> n) {
        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);
        addNode(node);
        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

import java.util.LinkedList

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


func treeInsert[T](tree: var DoublyLinkedNode[T]; node: DoublyLinkedNode[T]) =
  if tree.isNil: tree = node
  elif node.value < tree.value: tree.prev.treeInsert(node)
  else: tree.next.treeInsert(node)


func listFromTree[T](list: var DoublyLinkedList[T]; node: DoublyLinkedNode[T]) =
  if node.isNil: return
  let prev = node.prev
  let next = node.next
  list.listFromTree(prev)
  list.append(node)
  list.listFromTree(next)


func treeSort[T](list: DoublyLinkedList[T]): DoublyLinkedList[T] =
  var list = list
  if list.head == list.tail: return list
  var n = list.head
  var root: DoublyLinkedNode[T] = nil
  while not n.isNil:
    var next = n.next
    n.next = nil
    n.prev = nil
    root.treeInsert(n)
    n = next
  result = initDoublyLinkedList[T]()
  result.listFromTree(root)


randomize()
var list1 = initDoublyLinkedList[int]()
for i in 0..15: list1.append(rand(10..99))
echo "Before sort: ", list1
echo "After sort:  ", list1.treeSort()
echo()

var list2 = initDoublyLinkedList[string]()
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

-- doubly linked list:
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];
printLinkedList(@ll, "%d", False);
my @lls = treeSort(@ll);
printLinkedList(@lls, "%d", True);

my @ll2 = <d c e b a>;
#my @ll2 = <one two three four five six seven eight nine ten>;
printLinkedList(@ll2, "%s", False);
my @lls2 = treeSort(@ll2);
printLinkedList(@lls2, "%s", True);

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;
   fcn add(value){
      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)();
...