N-ary Tree Level Order Traversal
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0429.N-ary%20Tree%20Level%20Order%20Traversal/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0429.N-ary%20Tree%20Level%20Order%20Traversal/images/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>
Tree; Breadth-First Search
/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<Node> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int n = q.size(); n > 0; --n) {
                root = q.poll();
                t.add(root.val);
                q.addAll(root.children);
            }
            ans.add(t);
        }
        return ans;
    }
}