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Given the following tree structure in Haskell:

data Tree = Leaf Int | Node Int Tree Tree deriving Show

How can I get Haskell to return a list of the data in pre-order?

e.g. given a tree:

Node 1 (Leaf 2) (Leaf 3)

return something like:

preorder = [1,2,3]
Gravy
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3 Answers3

5

You could aim to a more general solution and make your data type an instance of Foldable. There is a very similar example at hackage, but that implements a post-order visit. If you want to support pre-order visits you will have to write something like this:

import qualified Data.Foldable as F

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

instance F.Foldable Tree where
    foldr f z (Leaf x) = f x z
    foldr f z (Node k l r) = f k (F.foldr f (F.foldr f z r) l)

With this, you'll be able to use every function that works on Foldable types, like elem, foldr, foldr, sum, minimum, maximum and such (see here for reference).

In particular, the list you are searching for can be obtain with toList. Here are some examples of what you could write by having that instance declaration:

*Main> let t = Node 1 (Node 2 (Leaf 3) (Leaf 4)) (Leaf 5)
*Main> F.toList t
[1,2,3,4,5]
*Main> F.foldl (\a x -> a ++ [x]) [] t
[1,2,3,4,5]
*Main> F.foldr (\x a -> a ++ [x]) [] t
[5,4,3,2,1]
*Main> F.sum t
15
*Main> F.elem 3 t
True
*Main> F.elem 12 t
False
Riccardo T.
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3

Use pattern matching

preorder (Leaf n) = [n]
preorder (Node n a b) = n:(preorder a) ++ (preorder b)
Philip JF
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  • I get the following error: Couldn't match expected type `[t0]' with actual type `Tree' In the pattern: Node n a b In an equation for `preorder' preorder (Node n a b) = n : (preorder a) ++ (preorder b) – Gravy Feb 29 '12 at 01:00
  • @Gravy That is bizarre. Try adding an explicit type signature `preorder :: Tree -> [Int]`. It doesn't make sense to me that you would be having an error. – Philip JF Feb 29 '12 at 02:10
  • That's weird I just tested it and seems to be working. Are you sure you copied everything correctly? – mck Feb 29 '12 at 02:14
  • @Gravy Also make sure you have `preorder (Leaf n) = [n]` not `preorder (Lead n) = n`. Those `[ ]` are important. – Philip JF Feb 29 '12 at 02:18
  • preorder(Leaf n) = [n] preorder(Node n treeL treeR) = [n] ++ preorder treeL ++ preorder treeR – Gravy Mar 19 '12 at 03:21
2

Ok, sorry about the late reply, but I got this working as follows:

preorder(Leaf n) = [n]
preorder(Node n treeL treeR) = [n] ++ preorder treeL ++ preorder treeR'code'

This however does not work for me still

preorder (Leaf n) = [n]   
preorder (Node n a b) = n:(preorder a) ++ (preorder b)
Gravy
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