Introduction

In this blog post, we will explore how to solve the LeetCode problem "102" 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 the level order traversal of its nodes' values. (i.e., from left to right, level by level). Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]] Example 2: Input: root = [1] Output: [[1]] Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 2000]. -1000 <= Node.val <= 1000

Explanation

Here's the solution to perform a level order (breadth-first) traversal of a binary tree:

• Key Idea: Use a queue to store nodes at each level. Process nodes level by level, adding their children to the queue for the next level.
• Iteration: The outer loop iterates while the queue is not empty (while we still have levels to process). The inner loop processes all nodes at the current level.

• Result: Store the values of nodes at each level in a separate list and append it to the final result list.

• Complexity: Time: O(N), where N is the number of nodes. Space: O(W), where W is the maximum width of the tree (worst case O(N) for a complete binary tree).

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

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

        while queue:
            level_size = len(queue)
            current_level = []

            for _ in range(level_size):
                node = queue.popleft()
                current_level.append(node.val)

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

            result.append(current_level)

        return result