Parsing 在guards haskell中使用大小写表达式时出现分析错误

我是Haskell的新手，语法有问题。我要做的是给定数据和此数据类型的树，找到树中相应节点的路径。我相信我的函数逻辑是正确的，但我不确定如何使它有效。我尝试过将选项卡更改为空格

-- | Binary trees with nodes labeled by values of an arbitrary type.
data Tree a
   = Node a (Tree a) (Tree a)
   | End
  deriving (Eq,Show)

-- | One step in a path, indicating whether to follow the left subtree (L)
--   or the right subtree (R).
data Step = L | R
  deriving (Eq,Show)

-- | A path is a sequence of steps. Each node in a binary tree can be
--   identified by a path, indicating how to move down the tree starting
--   from the root.
type Path = [Step]
pathTo :: Eq a => a -> Tree a -> Maybe Path
pathTo a End = Nothing
pathTo a (Node b l r)
    | a == b = Just []
    | case (pathTo a l) of
                Just p  -> Just [L:p]
                Nothing -> case (pathTo a r) of
                                    Just p  -> Just [R:p]
                                    Nothing -> Nothing

这就是错误：

  parse error (possibly incorrect indentation or mismatched brackets)

---

这里潜在的问题是，它看起来不像一个保护：保护是一个类型为

Bool

的表达式，这决定了保护是否“启动”。这里很可能是`否则：

pathTo :: Eq a => a -> Tree a -> Maybe Path
pathTo a End = Nothing
pathTo a (Node b l r)
    | a == b = Just []
    | otherwise = case (pathTo a l) of
                Just p  -> Just (L:p)
                Nothing -> case (pathTo a r) of
                                    Just p  -> Just (R:p)
                                    Nothing -> Nothing

---

这里

xy

如果是

x

，则取

x

，否则取

y

。我们使用

（L:）…

来预先包装在

Just

数据构造函数中的列表，或者返回

Nothing

，以防

是
Nothing
这看起来不像一个守卫，否则需要
否则=…
这看起来不像是一个解析错误，您的类型错误。您的编辑使答案与问题无关。如果您有后续问题，请发布新问题。
import Control.Applicative((<|>))

pathTo :: Eq a => a -> Tree a -> Maybe Path
pathTo a End = Nothing
pathTo a (Node b l r)
    | a == b = Just []
    | otherwise = ((L:) <$> pathTo a l) <|> ((R:) <$> pathTo a r)