Problem 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:

https://assets.leetcode.com/uploads/2019/05/03/capture.JPG

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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, 104].
  • 105 <= Node.val <= 105

Solution:

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# 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: Optional[TreeNode]) -> int:
        max, level, minLevel = float("-inf"), 0, 0
        queue = deque([root])
        
        while queue:
            level += 1
            sum = 0
            for i in range(len(queue)):
                front = queue.popleft()
                sum += front.val
                if front.left: queue.append(front.left)
                if front.right: queue.append(front.right)
            if sum > max:
                max = sum
                minLevel = level
        return minLevel

time complexity: $O()$
space complexity: $O()$
reference:
related problem: