★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ ?微信公众号:为敢(WeiGanTechnologies) ?博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)
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?微信公众号:为敢(WeiGanTechnologies)
?博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)
?GitHub地址:https://github.com/strengthen/LeetCode
?原文地址:https://www.cnblogs.com/strengthen/p/11371957.html
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Given the root
of a binary tree, the level of its root is 1
, the level of its children is 2
, and so on.
Return the smallest level X
such that the sum of all the values of nodes at level X
is maximal.
Example 1:
Input: [1,7,0,7,-8,null,null]
Output: 2 Explanation: Level 1 sum = 1. Level 2 sum = 7 + 0 = 7. Level 3 sum = 7 + -8 = -1. So we return the level with the maximum sum which is level 2.
Note:
- The number of nodes in the given tree is between
1
and10^4
. -10^5 <= node.val <= 10^5
给你一个二叉树的根节点 root
。设根节点位于二叉树的第 1
层,而根节点的子节点位于第 2
层,依此类推。
请你找出层内元素之和 最大 的那几层(可能只有一层)的层号,并返回其中 最小 的那个。
示例:
输入:[1,7,0,7,-8,null,null] 输出:2 解释: 第 1 层各元素之和为 1, 第 2 层各元素之和为 7 + 0 = 7, 第 3 层各元素之和为 7 + -8 = -1, 所以我们返回第 2 层的层号,它的层内元素之和最大。
提示:
- 树中的节点数介于
1
和10^4
之间 -10^5 <= node.val <= 10^5