Folding over Rose Tree

Question

Given the following Algrebaic Data Structure:

data Tree a = Node {
    rootLabel :: a,
    subForest :: [Tree a]
}  deriving (Show)

and fold :

treeFold :: (a -> [b] -> b) -> Tree a -> b
treeFold f (Node x ts) = f x (map (treeFold' f) ts)

How can I get an [a] from a Tree a ?

Answer 1

Do you mean using fold? You can get a function Tree a -> [a] pretty straightforward:

collapse :: Tree a -> [a]
collapse (Node x ts) = x : (concat $ map collapse ts)

Prelude> let t = Node 3 [Node 2 [], Node 4 [], Node 6 []]
Prelude> collapse t
[3,2,4,6]

If you specifically want to use fold , I guess you could do something similar:

collapse' :: Tree a -> [a]
collapse' = treeFold (\x tss -> x : (concat tss))

Prelude> collapse' t
[3,2,4,6]

I personally think the first version is clearer.