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;
    }
};