Haskell - Creating a list by using functions

Question

I want to do what the title says, but it keeps me showing errors. I put my code below.

data Tree a = Leaf a | Branch (Tree a) (Tree a) deriving Show

treeToList :: Tree a -> [a]
treeToList (Leaf x) = [x]
treeToList (Branch a b) = (treeToList a):(treeToList b)

It would show something like this:

treeToList Branch (Branch (Leaf 2) (Leaf 3)) (Leaf 4)

[2,3,4]

I see this kind of tree (with data in leaves but not in branches) every now and then, and always wonder what's the point of it. — n. m. could be an AI
– n. m. could be an AI, Commented Nov 17, 2014 at 18:55
@Sibi Consider this: data List a = Continue (List a) | Finish a. Makes sense? — n. m. could be an AI
– n. m. could be an AI, Commented Nov 17, 2014 at 19:02
@n.m. But your data structure only allows to store one data. (eg: Continue (Continue (Continue (Finish 3)))) ? — Sibi
– Sibi, Commented Nov 17, 2014 at 19:08
@n.m.- my filesystem is a tree with data in the leaves (files) but not in the branches (the dirs).... Makes sense to me. — jamshidh
– jamshidh, Commented Nov 17, 2014 at 19:12

Sibi · Accepted Answer · 2014-11-17 18:51:49Z

2

In the first pattern match you get a value of type list. Now all you have to do is concatenate them when you pattern match the Branch constructor.

treeToList :: Tree a -> [a]
treeToList (Leaf x) = [x]
treeToList (Branch a b) = (treeToList a) ++ (treeToList b)

Demo in ghci:

*Main> treeToList (Branch (Branch (Leaf 2) (Leaf 3)) (Leaf 4))
[2,3,4]

answered Nov 17, 2014 at 18:51

Sibi

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