Folding over a Rose Tree : Tree a -> [a]

Question

I posted this question on StackOverflow.

Given this algebraic data type:

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

and function:

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

Write a function (using treeFold) with the signature: Tree a -> [a].

Given this example:

ghci> stringTree
Node {rootLabel = "foo", 
      subForest = [Node {rootLabel = "bar",   subForest = []},
                   Node {rootLabel = "bippy", subForest = []},
                   Node {rootLabel = "baz", subForest = []} ]}

I came up with:

ghci> treeFold' (\x y -> x : (join y) ) stringTree
["foo","bar","bippy","baz"]

Please review this implementation.

I'm not sure if introducing join here (from Control.Monad) is overkill. Note that I don't have a good sense as to whether it's OK to casually use Monads - mainly using the right abstraction - not going overboard.

Answer 1

I don't think that using monads here itself is overkill, but I would suggest using concat instead of join.

There are no different Monads that could be an argument to join here. Using concat is more apropriate (imo).