Problem URL: Maximum level sum of a binary tree

We will traverse the tree with BFS and we will keep on iterating over the nodes of the tree. We will keep on adding the nodes of the current level to a queue and we will keep on adding the values of the nodes of the current level to a variable. We will keep on doing this until we have visited all the nodes of the tree. At the end, we will return the maximum sum of the nodes of the 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:
        res = 0
        level, _max = 1, -math.inf
        q = collections.deque([root])
        while q:
            qLen = len(q)
            levelSum = 0
            for _ in range(qLen):
                node = q.pop()
                levelSum += node.val
                if node.left: q.appendleft(node.left)
                if node.right: q.appendleft(node.right)
            if levelSum > _max:
                _max = levelSum
                res = level
            level += 1
        return res

Time complexity: O(n)
Space complexity: O(n)