Question

我有一棵这样的树：

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

我需要一个找到树顶的函数。我写了这个：

root :: Tree a -> a
root (Leaf a) = a
root (Node a _ _ _) = a

完美有效，但是当我有一棵空树时，我有一条错误信息。

如果我添加

root Empty = Empty

我再次出错，因为它没有返回a类型的值。我该怎么办？

Answer 1

在a构造函数的情况下，Empty没有合理的值。在这种情况下，类型Maybe完全符合您的需要：合理值或无值。因此，以下将是实现您的功能的惯用方法：

root :: Tree a -> Maybe a
root (Leaf a) = Just a
root (Node a _ _ _) = Just a
root _ = Nothing

您可以使用functors, applicative functors and monads轻松地将此功能与库的其他功能组合在一起。 E.g：

functionOnPlainA :: a -> a
functionOnPlainA = error "implement me"

-- So simply we can lift our plain function to work on values of type `Maybe`.
-- This is a Functor example.
functionOnMaybeA :: Maybe a -> Maybe a
functionOnMaybeA = fmap functionOnPlainA

-- Because the above is so simple, it's conventional to inline it,
-- as in the following:
applyFunctionOnPlainAToResultOfRoot :: Tree a -> Maybe a
applyFunctionOnPlainAToResultOfRoot tree = fmap functionOnPlainA $ root tree

-- And here is an example of how Monads can serve us:
applyRootToMaybeTree :: Maybe (Tree a) -> Maybe a
applyRootToMaybeTree maybeTree = maybeTree >>= root

在haskell中找到树的顶部（根）

1 个答案: