Re: [Haskell-cafe] help in tree folding

On Tue, May 6, 2008 at 8:20 AM, patrik osgnach wrote:

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Hi. I'm learning haskell but i'm stuck on a generic tree folding exercise. i must write a function of this type treefoldr::(Eq a,Show a)=>(a->b->c)->c->(c->b->b)->b->Tree a->c Tree has type data (Eq a,Show a)=>Tree a=Void | Node a [Tree a] deriving (Eq,Show) as an example treefoldr (:) [] (++) [] (Node '+' [Node '*' [Node 'x' [], Node 'y' []], Node 'z' []]) must return "+∗xyz" any help? (sorry for my bad english)

Having a (Tree a) parameter, where Tree is defined as an algebraic data type, also immediately suggests that you should do some pattern-matching to break treefoldr down into cases:

treefoldr f y g z Void = ? treefoldr f y g z (Node x t) = ?

-Brent so far i have tried

Brent Yorgey ha scritto: treefoldr f x g y Void = x treefoldr f x g y (Node a []) = f a y treefoldr f x g y (Node a (t:ts)) = treefoldr f x g (g (treefoldr f x g y t) y) (Node a ts) but it is incorrect. i can't figure out how to build the recursive call thanks for the answer Patrik