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

Updated
2 min read

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, 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

    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

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