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: 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

SOLUTION:

# 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 inorder(self, root, l):
        if root:
            self.inorder(root.left, l + 1)
            self.levels[l] = self.levels.get(l, 0) + root.val
            self.inorder(root.right, l + 1)

    def maxLevelSum(self, root: Optional[TreeNode]) -> int:
        self.levels = {}
        self.inorder(root, 1)
        return -max([(v, -k) for k, v in self.levels.items()])[1]