树的 Haskell map

Question

我的树由

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

我还声明了一个测试树。

myTree = Node (Node (Leaf 1) (Leaf 2)) (Leaf 3)

我想要做的是创建一个函数 maptree f 它将作用于 Leaf。更具体地说， f x = x +1 ,

然后 maptree f myTree将返回

Node (Node (Leaf 2) (Leaf 3)) (Leaf 4)

我的解决方案是

maptree f (Leaf a)= Leaf (f a)
maptree f (Node xl xr ) = Node (maptree xl) (maptree xr)

但它会返回以下错误

Couldn't match expected type `Tree a'
       against inferred type `Tree t -> Tree t'
Probable cause: `maptree' is applied to too few arguments
In the first argument of `Node', namely `(maptree xl)'
In the expression: Node (maptree xl) (maptree xr)

失败，加载模块:无。

但是，如果我这样做

maptree (Leaf a)= Leaf ( a + 1)
maptree (Node xl xr ) = Node (maptree xl) (maptree xr)

它确实有效。

我看不出第一个函数和第二个函数之间的区别。我如何得到错误？谢谢。

Answer 1

您缺少递归 maptree 调用中的函数:

maptree f (Leaf a)= Leaf (f a) 
maptree f (Node xl xr ) = Node (maptree xl) (maptree xr)

应该

maptree f (Leaf a)= Leaf (f a) 
maptree f (Node xl xr ) = Node (maptree f xl) (maptree f xr)