27 votes

Joli imprimé un arbre

Disons que j'ai une structure de données d'arbre binaire définie comme suit

 type 'a tree =
    | Node of 'a tree * 'a * 'a tree
    | Nil
 

J'ai une instance d'arbre comme suit:

 let x =
  Node
    (Node (Node (Nil,35,Node (Nil,40,Nil)),48,Node (Nil,52,Node (Nil,53,Nil))),
     80,Node (Node (Nil,82,Node (Nil,83,Nil)),92,Node (Nil,98,Nil)))
 

J'essaie d'imprimer l'arbre en quelque chose de facile à interpréter. De préférence, j'aimerais imprimer l'arborescence dans une fenêtre de console comme celle-ci:

         _______ 80 _______
       /                  \
    _ 48 _              _ 92 _
   /      \            /      \
 35       52         82       98
   \       \        /
    40      53    83
 

Quelle est la manière la plus simple de faire sortir mon arbre dans ce format?

27voto

Brian Points 82719

Si vous voulez qu'il soit très joli, vous pouvez voler à environ 25 lignes de code à partir de cette entrée de blog à dessiner avec WPF.

Mais je vais le code ascii solution peu trop, sans doute.

MODIFIER

Ok, wow, c'était dur.

Je ne suis pas certain qu'il est tout à fait correct, et je ne peux pas aider mais pense qu'il y a probablement une meilleure abstraction. Mais de toute façon... profitez-en!

(Voir à la fin du code pour un grand exemple qui est assez joli.)

type 'a tree =    
    | Node of 'a tree * 'a * 'a tree
    | Nil

(*
For any given tree
     ddd
     / \
   lll rrr  
we think about it as these three sections, left|middle|right (L|M|R):
     d | d | d
     / |   | \
   lll |   | rrr  
M is always exactly one character.
L will be as wide as either (d's width / 2) or L's width, whichever is more (and always at least one)
R will be as wide as either ((d's width - 1) / 2) or R's width, whichever is more (and always at least one)
     (above two lines mean 'dddd' of even length is slightly off-center left)
We want the '/' to appear directly above the rightmost character of the direct left child.
We want the '\' to appear directly above the leftmost character of the direct right child.
If the width of 'ddd' is not long enough to reach within 1 character of the slashes, we widen 'ddd' with
    underscore characters on that side until it is wide enough.
*)

// PrettyAndWidthInfo : 'a tree -> string[] * int * int * int
// strings are all the same width (space padded if needed)
// first int is that total width
// second int is the column the root node starts in
// third int is the column the root node ends in
// (assumes d.ToString() never returns empty string)
let rec PrettyAndWidthInfo t =
    match t with
    | Nil -> 
        [], 0, 0, 0
    | Node(Nil,d,Nil) -> 
        let s = d.ToString()
        [s], s.Length, 0, s.Length-1
    | Node(l,d,r) ->
        // compute info for string of this node's data
        let s = d.ToString()
        let sw = s.Length
        let swl = sw/2
        let swr = (sw-1)/2
        assert(swl+1+swr = sw)  
        // recurse
        let lp,lw,_,lc = PrettyAndWidthInfo l
        let rp,rw,rc,_ = PrettyAndWidthInfo r
        // account for absent subtrees
        let lw,lb = if lw=0 then 1," " else lw,"/"
        let rw,rb = if rw=0 then 1," " else rw,"\\"
        // compute full width of this tree
        let totalLeftWidth = (max (max lw swl) 1)
        let totalRightWidth = (max (max rw swr) 1)
        let w = totalLeftWidth + 1 + totalRightWidth
(*
A suggestive example:
     dddd | d | dddd__
        / |   |       \
      lll |   |       rr
          |   |      ...
          |   | rrrrrrrrrrr
     ----       ----           swl, swr (left/right string width (of this node) before any padding)
      ---       -----------    lw, rw   (left/right width (of subtree) before any padding)
     ----                      totalLeftWidth
                -----------    totalRightWidth
     ----   -   -----------    w (total width)
*)
        // get right column info that accounts for left side
        let rc2 = totalLeftWidth + 1 + rc
        // make left and right tree same height        
        let lp = if lp.Length < rp.Length then lp @ List.init (rp.Length-lp.Length) (fun _ -> "") else lp
        let rp = if rp.Length < lp.Length then rp @ List.init (lp.Length-rp.Length) (fun _ -> "") else rp
        // widen left and right trees if necessary (in case parent node is wider, and also to fix the 'added height')
        let lp = lp |> List.map (fun s -> if s.Length < totalLeftWidth then (nSpaces (totalLeftWidth - s.Length)) + s else s)
        let rp = rp |> List.map (fun s -> if s.Length < totalRightWidth then s + (nSpaces (totalRightWidth - s.Length)) else s)
        // first part of line1
        let line1 =
            if swl < lw - lc - 1 then
                (nSpaces (lc + 1)) + (nBars (lw - lc - swl)) + s
            else
                (nSpaces (totalLeftWidth - swl)) + s
        // line1 right bars
        let line1 =
            if rc2 > line1.Length then
                line1 + (nBars (rc2 - line1.Length))
            else
                line1
        // line1 right padding
        let line1 = line1 + (nSpaces (w - line1.Length))
        // first part of line2
        let line2 = (nSpaces (totalLeftWidth - lw + lc)) + lb 
        // pad rest of left half
        let line2 = line2 + (nSpaces (totalLeftWidth - line2.Length))
        // add right content
        let line2 = line2 + " " + (nSpaces rc) + rb
        // add right padding
        let line2 = line2 + (nSpaces (w - line2.Length))
        let resultLines = line1 :: line2 :: ((lp,rp) ||> List.map2 (fun l r -> l + " " + r))
        for x in resultLines do
            assert(x.Length = w)
        resultLines, w, lw-swl, totalLeftWidth+1+swr
and nSpaces n = 
    String.replicate n " "
and nBars n = 
    String.replicate n "_"

let PrettyPrint t =
    let sl,_,_,_ = PrettyAndWidthInfo t
    for s in sl do
        printfn "%s" s

let y = Node(Node (Node (Nil,35,Node (Node(Nil,1,Nil),88888888,Nil)),48,Node (Nil,777777777,Node (Nil,53,Nil))),     
             80,Node (Node (Nil,82,Node (Nil,83,Nil)),1111111111,Node (Nil,98,Nil)))
let z = Node(y,55555,y)
let x = Node(z,4444,y)

PrettyPrint x
(*
                                   ___________________________4444_________________
                                  /                                                \
                      ________55555________________                         ________80
                     /                             \                       /         \
            ________80                      ________80             _______48         1111111111
           /         \                     /         \            /        \            /  \
   _______48         1111111111    _______48         1111111111 35         777777777  82   98
  /        \            /  \      /        \            /  \      \             \       \
35         777777777  82   98   35         777777777  82   98     88888888      53      83
  \             \       \         \             \       \            /
  88888888      53      83        88888888      53      83           1
     /                               /
     1                               1
*)

4voto

Doug Currie Points 26016

Si cela ne vous dérange pas de tourner la tête sur le côté, vous pouvez d'abord imprimer la profondeur de l'arbre, un nœud sur une ligne, en passant récursivement la profondeur dans l'arbre et en imprimant depth*N espaces sur la ligne avant le nœud.

Voici le code Lua:

 tree={{{nil,35,{nil,40,nil}},48,{nil,52,{nil,53,nil}}},
      80,{{nil,82,{nil,83,nil}},92 {nil,98,nil}}}

function pptree (t,depth) 
  if t ~= nil
  then pptree(t[3], depth+1)
    print(string.format("%s%d",string.rep("  ",depth), t[2]))
    pptree(t[1], depth+1)
  end
end
 

Tester:

 > pptree(tree,4)
        98
      92
          83
        82
    80
          53
        52
      48
          40
        35
>
 

4voto

Isaac Kleinman Points 518

Cet article semble sympa http://llimllib.github.com/pymag-trees/

2voto

desco Points 12018

Peut-être que cela peut aider: dessiner des arbres en ML

1voto

Chip Uni Points 4739

Bien que ce ne soit pas exactement la bonne sortie, j'ai trouvé une réponse sur http://www.christiankissig.de/cms/files/ocaml99/problem67.ml :

 (* A string representation of binary trees

Somebody represents binary trees as strings of the following type (see example opposite):

a(b(d,e),c(,f(g,)))

a) Write a Prolog predicate which generates this string representation, if the tree 
is given as usual (as nil or t(X,L,R) term). Then write a predicate which does this 
inverse; i.e. given the string representation, construct the tree in the usual form. 
Finally, combine the two predicates in a single predicate tree_string/2 which can be 
used in both directions.

b) Write the same predicate tree_string/2 using difference lists and a single 
predicate tree_dlist/2 which does the conversion between a tree and a difference 
list in both directions.

For simplicity, suppose the information in the nodes is a single letter and there are 
no spaces in the string. 
*)

type bin_tree = 
	Leaf of string
|	Node of string * bin_tree * bin_tree
;;

let rec tree_to_string t =
    match t with
            Leaf s -> s
    |       Node (s,tl,tr) -> 
                    String.concat "" 
                            [s;"(";tree_to_string tl;",";tree_to_string tr;")"]
;;
 

Prograide.com

Prograide est une communauté de développeurs qui cherche à élargir la connaissance de la programmation au-delà de l'anglais.
Pour cela nous avons les plus grands doutes résolus en français et vous pouvez aussi poser vos propres questions ou résoudre celles des autres.

Powered by:

X