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@ckoparkar
Last active September 7, 2017 05:08
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Implementation of Binary Search Tree in Elm
{-----------------------------------------------------------------
A "Tree" represents a binary tree. A "Node" in a binary tree
always has two children. A tree can also be "Empty". Below I have
defined "Tree" and a number of useful functions.
This example also includes some challenge problems :)
-----------------------------------------------------------------}
import Graphics.Element exposing (..)
import Text
type Tree a
= Empty
| Node a (Tree a) (Tree a)
empty : Tree a
empty =
Empty
singleton : a -> Tree a
singleton v =
Node v Empty Empty
insert : comparable -> Tree comparable -> Tree comparable
insert x tree =
case tree of
Empty ->
singleton x
Node y left right ->
if x > y then
Node y left (insert x right)
else if x < y then
Node y (insert x left) right
else
tree
fromList : List comparable -> Tree comparable
fromList xs =
List.foldl insert empty xs
depth : Tree a -> Int
depth tree =
case tree of
Empty -> 0
Node v left right ->
1 + max (depth left) (depth right)
map : (a -> b) -> Tree a -> Tree b
map f tree =
case tree of
Empty -> Empty
Node v left right ->
Node (f v) (map f left) (map f right)
t1 = fromList [1,2,3]
t2 = fromList [2,7,5]
main : Element
main =
flow down
[ display "depth" depth t1
, display "depth" depth t2
, display "map ((+)1)" (map ((+)1)) t2
, display "sum" sum t2
, display "flatten" flatten t2
, display "isElement" (isElement 7) t2
, display "sum with fold" (fold (+) 0) t2
, display "flatten with fold" (fold (::) []) t2
--, display "isElement with fold" (fold ((==) >> (||)) False) t2
, display "map with fold" (map' ((+)1)) t2
, display "depth with fold" depth' t2
]
map' : (a -> comparable) -> Tree a -> Tree comparable
map' f tree = fold (\x xs -> insert (f x) xs) empty tree
depth' : Tree a -> Int
depth' = fold (\x n -> 1 + n) 0
display : String -> (Tree a -> b) -> Tree a -> Element
display name f value =
name ++ " (" ++ toString value ++ ") &rArr;\n " ++ toString (f value) ++ "\n "
|> Text.fromString
|> Text.monospace
|> leftAligned
sum : Tree number -> number
sum tree =
case tree of
Empty -> 0
Node x left right ->
x + sum left + sum right
flatten : Tree a -> List a
flatten tree =
case tree of
Empty -> []
Node x left right ->
flatten left ++ [x] ++ flatten right
isElement : comparable -> Tree comparable -> Bool
isElement a tree =
case tree of
Empty -> False
Node x left right ->
if x == a then
True
else if a > x then
isElement a right
else
isElement a left
fold : (a -> b -> b) -> b -> Tree a -> b
fold fn base tree =
let
xs = flatten tree
fold' f v ys =
case ys of
[] -> base
(y::ys) -> f y (fold' f v ys)
in
fold' fn base xs
{-----------------------------------------------------------------
Exercises:
(1) Sum all of the elements of a tree.
sum : Tree number -> number
(2) Flatten a tree into a list.
flatten : Tree a -> List a
(3) Check to see if an element is in a given tree.
isElement : a -> Tree a -> Bool
(4) Write a general fold function that acts on trees. The fold
function does not need to guarantee a particular order of
traversal.
fold : (a -> b -> b) -> b -> Tree a -> b
(5) Use "fold" to do exercises 1-3 in one line each. The best
readable versions I have come up have the following length
in characters including spaces and function name:
sum: 16
flatten: 21
isElement: 46
See if you can match or beat me! Don't forget about currying
and partial application!
(6) Can "fold" be used to implement "map" or "depth"?
(7) Try experimenting with different ways to traverse a
tree: pre-order, in-order, post-order, depth-first, etc.
More info at: http://en.wikipedia.org/wiki/Tree_traversal
-----------------------------------------------------------------}
@pakx
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pakx commented May 13, 2016

@avinashsmiles, the following comes without testing, but consider:
your code: fold f (fold f (f b n) left) right
should be: fold f (fold f (f n b) left) right

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