# Solving Leetcode Interviews in Seconds with AI: N-ary Tree Level Order Traversal


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "429" 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 an n-ary tree, return the level order traversal of its nodes' values. Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).   Example 1:   Input: root = [1,null,3,2,4,null,5,6] Output: [[1],[3,2,4],[5,6]]  Example 2:   Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14] Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]    Constraints:  The height of the n-ary tree is less than or equal to 1000 The total number of nodes is between [0, 104]  

	# Explanation
	Here's a breakdown of the approach, complexity, and the Python code for the n-ary tree level order traversal:

*   **Key Approach:**
    *   Utilize a queue to perform a Breadth-First Search (BFS) traversal of the n-ary tree.
    *   Process nodes level by level, adding children to the queue for the next level.
    *   Use a `while` loop and the `len(queue)` at the start of each level's iteration to delineate each level's node processing.

*   **Complexity:**
    *   Runtime Complexity: O(N), where N is the number of nodes in the n-ary tree.
    *   Storage Complexity: O(W), where W is the maximum width of the n-ary tree. In the worst-case scenario (a complete n-ary tree), W can be proportional to N.

	
	# Code
	```python
	"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

from collections import deque

class Solution:
    def levelOrder(self, root: 'Node') -> list[list[int]]:
        """
        Performs a level order traversal of an n-ary tree.

        Args:
            root: The root of the n-ary tree.

        Returns:
            A list of lists, where each inner list contains the node values at each level.
        """

        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.children:
                    for child in node.children:
                        queue.append(child)

            result.append(current_level)

        return result
	```