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: root = [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.
Example 2:
Input: root = [989,null,10250,98693,-89388,null,null,null,-32127] Output: 2
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. -105 <= Node.val <= 105
最大层内元素和。
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。
请返回层内元素之和 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。
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/maximum-level-sum-of-a-binary-tree
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这道题不难，属于树的BFS遍历。注意记录一个层数 level，如果当前层的节点值的和是最大的，不但要记录这个最大的和是多少，同时要记录一下这是第几层。
时间O(n)
空间O(n)
Java实现
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 public int maxLevelSum(TreeNode root) { 18 int res = Integer.MIN_VALUE; 19 int level = 1; 20 int maxLevel = 1; 21 Queue<TreeNode> queue = new LinkedList<>(); 22 queue.offer(root); 23 while (!queue.isEmpty()) { 24 int size = queue.size(); 25 int sum = 0; 26 for (int i = 0; i < size; i++) { 27 TreeNode cur = queue.poll(); 28 sum += cur.val; 29 if (cur.left != null) { 30 queue.offer(cur.left); 31 } 32 if (cur.right != null) { 33 queue.offer(cur.right); 34 } 35 } 36 if (sum > res) { 37 maxLevel = level; 38 res = sum; 39 } 40 level++; 41 } 42 return maxLevel; 43 } 44 }