题目地址：https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/
题目描述
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:
1、 Thenumberofnodesinthegiventreeisbetween1and10^4.；
2、 -10^5<=node.val<=10^5；
题目大意
二叉树中一层的节点和最大的时候的最小层号。
解题方法
BFS
这个题考的是层次遍历，可以有两种做法，分别是BFS和DFS，类似题目是102. Binary Tree Level Order Traversalopen in new window。这里使用的是BFS。
BFS需要一个队列存放当前层的所有叶子节点，然后出队列并且对这一层的所有叶子节点求和。
题目要求的是最大的和出现的最小层号，所以做个判断，如果当前层的和大于之前层，那么修改结果的层号。
C++代码如下：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxLevelSum(TreeNode* root) {
int res_sum = INT_MIN;
int res_level = 1;
queue<TreeNode*> que;
que.push(root);
int level = 1;
while (!que.empty()) {
int size = que.size();
int level_sum = 0;
while (size --) {
TreeNode* cur = que.front(); que.pop();
if (!cur) continue;
level_sum += cur->val;
que.push(cur->left);
que.push(cur->right);
}
if (level_sum > res_sum) {
res_sum = level_sum;
res_level = level;
}
level ++;
}
return res_level;
}
};
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