Maximum level sum of a binary tree

# 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:
        if not root:
            return 1
        q = []
        q.append(root)
        res = 0
        level = 1
        maxval = -math.inf
        while q:
            size = len(q)
            tmp = 0
            for i in range(size):
                node = q.pop(0)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
                tmp += node.val
            # print(level, tmp)
            if tmp > maxval:
                maxval = tmp
                res = level
            level += 1


        return res

Maximum Level Sum of a Binary Tree

Difficulty: 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: 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:

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