Formatted question description: https://leetcode.ca/all/1161.html

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/

// 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;
    }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxLevelSum(TreeNode root) {
        if (root == null)
            return 0;
        int maxSum = 0;
        int maxSumLevel = 0;
        int level = 0;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            level++;
            int curSum = 0;
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                curSum += node.val;
                TreeNode left = node.left, right = node.right;
                if (left != null)
                    queue.offer(left);
                if (right != null)
                    queue.offer(right);
            }
            if (curSum > maxSum) {
                maxSum = curSum;
                maxSumLevel = level;
            }
        }
        return maxSumLevel;
    }
}

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