1161. Maximum Level Sum of a Binary Tree

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1161. Maximum Level Sum of a Binary Tree

Description

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:

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Input: root = [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.

Example 2:

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Input: root = [989,null,10250,98693,-89388,null,null,null,-32127]
Output: 2

Constraints:

  • The number of nodes in the tree is in the range [1, 10^4].
  • -10^5 <= Node.val <= 10^5

Hints/Notes

  • binary tree

Solution

Language: C++

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/**
 * 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 maxVal = INT_MIN;
        int minLvl = -1;
        queue<TreeNode*> q;
        if (root)
            q.push(root);
        int level = 1;
        while (!q.empty()) {
            int n = q.size(), sum = 0;
            for (int i = 0; i < n; i++) {
                TreeNode* cur = q.front();
                q.pop();
                sum += cur->val;
                if (cur->left)
                    q.push(cur->left);
                if (cur->right)
                    q.push(cur->right);
            }
            if (sum > maxVal) {
                maxVal = sum;
                minLvl = level;
            }
            level++;
        }
        return minLvl;
    }
};