题目描述
思路
用BFS层序遍历来写。当前队列个数就是当前层的节点个数。再压入下一层节点的同时，计算当前层内元素和。
AC代码
/**
* 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 ans = -1e9 + 7, pos = 1;
queue<TreeNode*> q;
q.push(root);
int level = 0;
while(!q.empty()) {
level++;
int size = q.size(), sum = 0;
// 一层一层计算
while(size -- > 0) {
TreeNode* cur = q.front(); q.pop();
sum += cur->val;
if (cur->left) {
q.push(cur->left);
}
if (cur->right) {
q.push(cur->right);
}
}
if(ans < sum) {
ans = sum;
pos = level;
}
}
return pos;
}
};