从树中删除空节点

Question

我想实现一个删除树中所有空子节点的函数：

makeUnhollow (Node 5 Empty Empty) => Leaf 5
makeUnhollow (Leaf 5) => Leaf 5
makeUnhollow (Node 5 (Leaf 4) Empty) => (Node 5 (Leaf 4) Empty)

这是我当前的代码：

makeUnhollow :: Tree a -> Tree a
makeUnhollow  (Node a Empty Empty)= Leaf a
makeUnhollow (Leaf a) = Leaf a
makeUnhollow a = a

但不知何故，我在这段代码中遇到了失败：

Tests.hs:130:
wrong result
expected: Node 6 (Leaf 5) (Leaf 7)
but got: Node 6 (Node 5 Empty Empty) (Node 7 Empty Empty)

Answer 1

删除所有 Empty 树中的孩子似乎有点困难：

• Node a Empty Empty可以变成Leaf a
• 如果你的根节点是Empty怎么办？
• Node a Empty (Leaf b) 会是什么？

我从您的测试中了解到，您的目标只是将Node a Empty Empty 变成Leaf a，而不在乎只有一个孩子是Empty。 Mark Seemann 建议将makeUnhollow 转换为递归函数，这意味着您必须让它在最后一种情况下调用自己：

makeUnhollow :: Tree a -> Tree a
makeUnhollow (Node a Empty Empty) = Leaf a
makeUnhollow (Leaf a) = Leaf a
makeUnhollow Empty = ?                --  don't forget to match Empty
makeUnhollow (Node a left right) = ?  --  recurse on left, right :: Tree a

我可以直接调用函数unhollow，因为这也是一个祈使动词。