# Solving Leetcode Interviews in Seconds with AI: Maximum Level Sum of a Binary Tree


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1161" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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.   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  

	# Explanation
	Here's the solution:

*   **Level-Order Traversal:** Use a breadth-first search (BFS) algorithm to traverse the binary tree level by level.
*   **Sum Calculation:** For each level, calculate the sum of the values of all nodes at that level.
*   **Maximum Sum Tracking:** Maintain variables to track the maximum level sum encountered so far and the corresponding level number.

*   **Runtime Complexity:** O(N), where N is the number of nodes in the binary tree.
*   **Storage Complexity:** O(W), where W is the maximum width of the binary tree (in the worst case, it could be O(N) for a complete binary tree).

	
	# Code
	```python
	from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def maxLevelSum(root: TreeNode) -> int:
    if not root:
        return 0

    queue = deque([root])
    max_sum = float('-inf')
    max_level = 0
    level = 0

    while queue:
        level += 1
        level_sum = 0
        level_size = len(queue)

        for _ in range(level_size):
            node = queue.popleft()
            level_sum += node.val

            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)

        if level_sum > max_sum:
            max_sum = level_sum
            max_level = level

    return max_level
	```