1116. Maximum Level Sum of a Binary Tree

• Difficulty: Medium

• Topics: Graph

• Similar Questions:

Problem:

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

Solutions:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxLevelSum(TreeNode* root) {
        if (root == nullptr)    return 0;
        unordered_map<int, int> sums;
        helper(root, 1, sums);

        int maxSum = INT_MIN;
        int maxLevel = -1;

        for (auto& entry : sums) {
            if (entry.second > maxSum) {
                maxSum = entry.second;
                maxLevel = entry.first;
            } else if (entry.second == maxSum && entry.first < maxLevel) {
                maxLevel = entry.first;
            }
        }

        return maxLevel;

    }
private:
    void helper(TreeNode* root, int level, unordered_map<int, int>& sums) {
        sums[level] += root->val;
        if (root->left) {
            helper(root->left, level + 1, sums);
        }
        if (root->right) {
            helper(root->right, level + 1, sums);
        }
    }

};