一、题目

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

二、题解

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:int maxLevelSum(TreeNode* root) {int res = 1;int maxSum = INT_MIN;queue<TreeNode*> q;q.push(root);int level = 0;while(!q.empty()){int size = q.size();int sum = 0;level++;while(size--){TreeNode* t = q.front();q.pop();sum += t->val;if(t->left) q.push(t->left);if(t->right) q.push(t->right);}if(sum > maxSum){maxSum = sum;res = level;}}return res;}
};