Solution 1
- TimeO(n)
- SpaceO(w)
Where, n is number of nodes in the tree, w is maximum width of the tree.
Simple BFS level order traversal problem.
Python · 2026-01-06
# 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:
'''
Time Complexity: O(n)
Space Complexity: O(w)
Where, n is number of nodes in the tree, w is maximum width of the tree.
Simple BFS level order traversal problem.
'''
def maxLevelSum(self, root: Optional[TreeNode]) -> int:
maxValue = root.val
maxLevel = 1
level = 1
nodes = [root]
while nodes:
childNodes = []
value = 0
for node in nodes:
if node.left:
childNodes.append(node.left)
if node.right:
childNodes.append(node.right)
value += node.val
if value > maxValue:
maxLevel = level
maxValue = value
nodes = childNodes
level += 1
return maxLevel