README.md

1161. Maximum Level Sum of a Binary Tree

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 (Python)

1. BFS

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxLevelSum(self, root: TreeNode) -> int:
        curr_level = [root]
        max_sum = root.val
        x = 1
        ret = 1

        while curr_level:
            curr_sum = sum(n.val for n in curr_level)
            if curr_sum > max_sum:
                max_sum = curr_sum
                ret = x

            curr_level = [c for n in curr_level for c in [n.left, n.right] if c]
            x += 1

        return ret