<p>Given an n-ary tree, return the <em>level order</em> traversal of its nodes&#39; values.</p>

<p><em>Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img src="https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png" style="width: 100%; max-width: 300px;" /></p>

<pre>
<strong>Input:</strong> root = [1,null,3,2,4,null,5,6]
<strong>Output:</strong> [[1],[3,2,4],[5,6]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png" style="width: 296px; height: 241px;" /></p>

<pre>
<strong>Input:</strong> 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]
<strong>Output:</strong> [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The height of the n-ary tree is less than or equal to <code>1000</code></li>
	<li>The total number of nodes is between <code>[0, 10<sup>4</sup>]</code></li>
</ul>

"""
# Definition for a Node.
class Node:
    def __init__(self, val: Optional[int] = None, children: Optional[List['Node']] = None):
        self.val = val
        self.children = children
"""

class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        if not root:
            return []
        
        queue = deque([root])
        result = []
        while queue:
            level = []
            for i in range(len(queue)):
                node = queue.popleft()
                level.append(node.val)
                if node.children:
                    for child in node.children:
                        queue.append(child)
                        
            result.append(level)
            
        return result
18200