Haskell：镜像二叉树错误：“无法匹配预期类型”

Question

所以我试图设计一个函数来返回一个二进制树，它是最初发送的二叉树的镜像。我想我已经差不多完成但是我得到一个奇怪的“无法匹配预期的类型。这是我的代码：

mirror :: BinTree a -> BinTree a    
mirror (Node x tL tR) = Node x (mirror tR) (mirror tL)

以下是错误：

hw1.hs:84:30:
    Couldn't match expected type `a' with actual type `BinTree a'
      `a' is a rigid type variable bound by
          the type signature for mirror :: BinTree a -> BinTree a
          at hw1.hs:83:11
    In the first argument of `Node', namely `x'
    In the expression: Node x (mirror tR) (mirror tL)
    In an equation for `mirror':
        mirror (Node tL x tR) = Node x (mirror tR) (mirror tL)

hw1.hs:84:33:
    Couldn't match expected type `a' with actual type `BinTree a'
      `a' is a rigid type variable bound by
          the type signature for mirror :: BinTree a -> BinTree a
          at hw1.hs:83:11
    In the return type of a call of `mirror'
    In the second argument of `Node', namely `(mirror tR)'
    In the expression: Node x (mirror tR) (mirror tL)
Failed, modules loaded: none.

这是我的错。我对树的定义不同。

data BinTree a = Empty | Node (BinTree a) a (BinTree a) deriving (Eq,Show)

* 新功能应该是：*

mirror :: BinTree a -> BinTree a
mirror Empty = Empty
mirror (Node tL x tR) = Node (mirror tR) x (mirror tL)

Answer 1

在@Davorak的建议中，我正在将我的评论转化为答案，即使它没有真正回答这个问题，只是解决了这个问题。

---

使用定义时

data BinTree a = Empty | Node a (BinTree a) (BinTree a) deriving (Eq)

OP的原始代码编译，但没有编译，因为他使用的定义是

data BinTree a = Empty | Node (BinTree a) a (BinTree a) deriving (Eq)

通过将函数mirror修改为BinTree a的正确定义的模式匹配，问题得以解决。