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, 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