Max Level Sum of A Binary Tree

Problem Statement:
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.

Approach

This was a problem which had some quite interesting solving... In this approach I used a BFS. I took one row at a time found its sum and compared it with the previous sum. If its greater than the previous one then I replaced the max Level with the current Level. Finally at the end of the looping I returned the max Level.

Algo Images

Code

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:
        if root == None: return None

        currentLevel: int = 1
        maxLevel: int = None
        maxSum: int = None

        BFSQueue: list[TreeNode] = [root]

        while len(BFSQueue) > 0:
            currentLength: int = len(BFSQueue)
            currentSum: int = 0

            for i in range(currentLength):
                currentNode: TreeNode = BFSQueue.pop(0)
                currentSum += currentNode.val

                if currentNode.left != None: BFSQueue.append(currentNode.left)
                if currentNode.right != None: BFSQueue.append(currentNode.right)

            if (maxSum == None) or (maxSum < currentSum):
                maxSum = currentSum
                maxLevel = currentLevel

            currentLevel += 1

        return maxLevel

Result Image

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