<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>

class Solution {
 public:
	vector<vector<int>> levelOrder(Node* root)
	{
		vector<vector<int>> res;
		if (!root)
		{
			return {};
		}

		queue<Node*> q;
		q.push(root);

		while (!q.empty())
		{
			int size = q.size();
            vector<int> level;
            for(int i = 0; i < size; ++i)
			{
				level.push_back(q.front()->val);

				for(auto child : q.front()->children)
				{
					q.push(child);
				}

				q.pop();
			}

			res.push_back(std::move(level));
		}

		return res;
	}
};

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