Pretty print a tree - f#

Question

Let's say I have a binary tree data structure defined as follows
type 'a tree =
| Node of 'a tree * 'a * 'a tree
| Nil
I have an instance of a tree as follows:
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)))
I'm trying to pretty-print the tree into something easy to interpret. Preferably, I'd like to print the tree in a console window like this:
_______ 80 _______
/ \
_ 48 _ _ 92 _
/ \ / \
35 52 82 98
\ \ /
40 53 83
What's an easy way to get my tree to output in that format?

Answer 1

If you want it to be very pretty, you could steal about 25 lines of code from this blog entry to draw it with WPF.
But I'll code up an ascii solution shortly too, probably.
EDIT
Ok, wow, that was hard.
I'm not certain it's entirely correct, and I can't help but think there's probably a better abstraction. But anyway... enjoy!
(See the end of the code for a large example that is rather pretty.)
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
*)

Answer 2

If you don't mind turning your head sideways, you can print the tree depth first, one node to a line, recursively passing the depth down the tree, and printing depth*N spaces on the line before the node.
Here's Lua code:
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
Test:
> pptree(tree,4)
98
92
83
82
80
53
52
48
40
35
>

Answer 3

Maybe this can help: Drawing Trees in ML

Answer 4

Although it's not exactly the right output, I found an answer at 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;")"]
;;

Answer 5

This is an intuition, I'm sure someone like Knuth had the idea, I'm too lazy
to check.
If you look at your tree as an one dimensional structure you will get an array
(or vector) of length L
This is easy to build with an "in order" recursive tree traversal: left,root,right
some calculations must be done to fill the gaps when the tree is unbalanced
2 dimension
_______ 80 _______
/ \
_ 48 _ _ 92 _
/ \ / \
35 52 82 98
\ \ /
40 53 83
1 dimension :
35 40 48 52 53 80 83 82 92 98
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
The pretty printed tree can be build using this array
(maybe with something recursive)
first using values at L/2 position, the X position is the
L/2 value * the default length (here it is 2 characters)
80
then (L/2) - (L/4) and (L/2) + (L/4)
48 92
then L/2-L/4-L/8, L/2-L/4+L/8, L/2+L/4-L/8 and L/2+L/4+L/8
35 52 82 98
...
Adding pretty branches will cause more positional arithmetics but it's trivial here
You can concatenate values in a string instead using an array, concatenation will
de facto calculate the best X postion and will allow different value size,
making a more compact tree.
In this case you will have to count the words in the string to extract
the values. ex: for the first element using the L/2th word of the string instead
of the L/2 element of the array. The X position in the string is the same in the tree.
N 35 40 48 N 52 53 80 83 82 N 92 N 98 N
80
48 92
35 52 82 98
40 53 83