[Leetcode] 107. 二叉树的层次遍历 II - V2EX
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Acceml
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[Leetcode] 107. 二叉树的层次遍历 II

  •  
  •   Acceml
    Acceml 2019-03-15 08:51:04 +08:00 2131 次点击
    这是一个创建于 2404 天前的主题,其中的信息可能已经有所发展或是发生改变。

    题目

    给定一个二叉树,返回其节点值自底向上的层次遍历。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)

    例如: 给定二叉树 [3,9,20,null,null,15,7],

     3 / \ 9 20 / \ 15 7 

    返回其自底向上的层次遍历为:

    [ [15,7], [9,20], [3] ] 

    题解

    利用层次遍历,层次遍历的时候进入下一层的时候记录一下当前队列中有几个元素。

    class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { List<List<Integer>> res = new LinkedList<>(); if (root == null) { return res; } LinkedList<TreeNode> queue = new LinkedList<>(); queue.add(root); while (!queue.isEmpty()) { int size = queue.size(); List<Integer> levelVal = new LinkedList<>(); while (size > 0) { TreeNode current = queue.poll(); if (current.left != null) { queue.add(current.left); } if (current.right != null) { queue.add(current.right); } levelVal.add(current.val); size--; } res.add(0, levelVal); } return res; } } 

    用递归去做。 用递归去做的关键在于需要把层数也带上。

    class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { List<List<Integer>> res = new LinkedList<>(); if (root == null) { return res; } helper(root, res, 0); return res; } public void helper(TreeNode root, List<List<Integer>> res, int depth) { if (root == null) { return; } if (depth == res.size()) { res.add(0, new LinkedList<>()); } List<Integer> current = res.get(res.size() - depth - 1); helper(root.left, res, depth + 1); helper(root.right, res, depth + 1); current.add(root.val); } } 

    热门阅读

    Leetcode 名企之路 手撕代码 QQ 群:805423079, 群密码:1024

    snowonion
        1
    snowonion  
       2019-03-15 11:01:18 +08:00
    https://www.codewars.com/kata/sort-binary-tree-by-levels 的 Haskell 解法的高票答案,稍加修改就能用在这里。

    (剧透警告)

    注意这里二叉树的定义方式是

    ```haskell
    data TreeNode a = TreeNode {
    left :: Maybe (TreeNode a),
    right :: Maybe (TreeNode a),
    value :: a
    } deriving Show
    ```

    解答:

    ```haskell
    levelOrderBottomUpHierarchical :: Maybe (TreeNode a) -> [[a]]
    levelOrderBottomUpHierarchical = reverse . takeWhile (not . null) . go
    where
    go Nothing = repeat []
    go (Just x) = [value x] : zipWith (++) (go $ left x) (go $ right x)
    ```

    测试:

    ```haskell
    leaf v = Just (TreeNode {left = Nothing, right = Nothing, value = v})

    t3 = Just (TreeNode {
    left = leaf 9,
    right = Just (TreeNode {
    left = leaf 15,
    right = leaf 7,
    value = 20
    }),
    value = 3
    })

    -- ghci 执行
    -- > levelOrderBottomUpHierarchical t3
    -- [[15,7],[9,20],[3]]
    ```
    snowonion
        2
    snowonion  
       2019-03-15 11:03:13 +08:00
    @snowonion #1
    缩进没了?! wtf
    msg7086
        3
    msg7086  
       2019-03-15 11:07:28 +08:00   1
    @snowonion 要贴 Markdown 需要开主题。
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