Introduction

In this blog post, we will explore how to solve the LeetCode problem "515" 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, return an array of the largest value in each row of the tree (0-indexed). Example 1: Input: root = [1,3,2,5,3,null,9] Output: [1,3,9] Example 2: Input: root = [1,2,3] Output: [1,3] Constraints: The number of nodes in the tree will be in the range [0, 104]. -231 <= Node.val <= 231 - 1

Explanation

Here's the approach to solve this problem efficiently:

• Level-Order Traversal: Perform a level-order (breadth-first) traversal of the binary tree to visit nodes row by row.
• Maximum Value Tracking: For each row, keep track of the maximum value encountered so far.

• Result Accumulation: Append the maximum value of each row to the result array.

• Runtime Complexity: O(N), where N is the number of nodes in the tree.

• Storage Complexity: O(W), where W is the maximum width of the tree (maximum number of nodes at any level). In the worst case, where the tree is a complete binary tree, W can be approximately N/2, hence O(N).

Code

    from collections import deque
from typing import List, Optional

# 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 largestValues(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []

        result = []
        queue = deque([root])

        while queue:
            level_size = len(queue)
            max_val = float('-inf')

            for _ in range(level_size):
                node = queue.popleft()
                max_val = max(max_val, node.val)

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

            result.append(max_val)

        return result