Haskell-为树类型创建折叠功能

Question

I have this simple data Tree : 我有这个简单的数据树：

data Tree = Leaf Int | Node [Tree]

And I have to devellop a fold function for this type : 我必须为此类型设计一个折叠函数：

foldTree :: (Int -> a) -> ([a] -> a) -> Tree -> a

for example : 例如：

foldTree (+1) sum (Node[ (Leaf 2), (Leaf 3)])

will return (2+1)+(3+1) = 7 For leafs, I have : 将返回（2 + 1）+（3 + 1）= 7对于叶子，我有：

foldTree f g (Leaf n)  = (f n)

But I have no ideas for develop the nodes's case. 但是我没有开发节点案例的想法。

I'm french, also sorry for the mistakes. 我是法国人，也为自己的错误感到抱歉。

Answer 1

It sometimes helps to look at what is available in scope and their types. 有时有助于查看范围及其类型中可用的内容。 Here's a solution: 这是一个解决方案：

foldTree f g (Leaf n)  = (f n)
foldTree f g (Node subtrees)  = 
  let as = map (foldTree f g) subtrees -- as :: [a]
  in g as