如何检查节点是否为二叉树中的叶子（Haskell）

由花自毁发布于 2020-05-24 06:32:09 haskellbinarybinary-tree

我需要检查节点是否是我的二叉树中的叶子。这是我当前的代码。

isLeaf :: Eq a => a -> BTree a -> Bool
isLeaf node tree@(Node el left right) 
    | tree == Empty                                 = False
    | right == Empty && left == Empty && node == el = True
    | otherwise                                     = isLeaf node left && isLeaf node right

它向我发送一条错误消息，提示： “ hw3_71937.hs：C：\ Users \ lenovo \ Desktop \ ... \\ HASKELL \ hw3_71937.hs：（22,1）-（25,91）：函数isLeaf中的非穷举模式”

我不知道如何递归检查下一个节点（如果它是叶子）。任何帮助将不胜感激。

续写呐份情 2020-05-24 06:32:09 回复
守卫：
```
  | tree == Empty = False
```
永远不可能是真的，因为在您的条款的开头：
```
isLeaf node tree@(Node el left right)
```
you say that this is a Node, not an Empty. This is also the case that you did not cover: in case you pass an Empty (and eventually you can pass an Empty due to recursion), no pattern will match.

话虽如此，我认为您使这一点变得太复杂了。您可以这样写：
```
isLeaf :: Eq a => a -> BTree a -> Bool
isLeaf _ Empty = False
isLeaf x (Node x' Empty Empty) = x == x'
isLeaf x (Node _ l r) = isLeaf x l || isLeaf x r
```
or with a where clause:
```
isLeaf :: Eq a => a -> BTree a -> Bool
isLeaf x = go
    where go Empty = False
          go (Node x' Empty Empty) = x == x'
          go (Node _ l r) = go l || go r
```