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1161. Maximum Level Sum of a Binary Tree (Medium)

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. The number of nodes in the given tree is between 1 and 10^4.
2. -10^5 <= node.val <= 10^5

Companies:
Google

Related Topics:
Graph

Solution 1. Level-order Traversal

// OJ: https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int maxLevelSum(TreeNode* root) {
        if (!root) return 0;
        int maxSum = 0, ans = 0, lv = 1;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            int cnt = q.size(), sum = 0;
            while (cnt--) {
                auto node = q.front();
                q.pop();
                sum += node->val;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            if (sum > maxSum) {
                maxSum = sum;
                ans = lv;
            }
            ++lv;
        }
        return ans;
    }
};