N-ary Tree Postorder Traversal
<p>Given the <code>root</code> of an n-ary tree, return <em>the postorder traversal of its nodes&#39; values</em>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0590.N-ary%20Tree%20Postorder%20Traversal/images/narytreeexample.png" style="width: 100%; max-width: 300px;" />
<pre>
<strong>Input:</strong> root = [1,null,3,2,4,null,5,6]
<strong>Output:</strong> [5,6,3,2,4,1]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0590.N-ary%20Tree%20Postorder%20Traversal/images/sample_4_964.png" style="width: 296px; height: 241px;" />
<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> [2,6,14,11,7,3,12,8,4,13,9,10,5,1]
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
	<li>The height of the n-ary tree is less than or equal to <code>1000</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Recursive solution is trivial, could you do it iteratively?</p>
Stack; Tree; Depth-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 {
    private List<Integer> ans = new ArrayList<>();

    public List<Integer> postorder(Node root) {
        dfs(root);
        return ans;
    }

    private void dfs(Node root) {
        if (root == null) {
            return;
        }
        for (Node child : root.children) {
            dfs(child);
        }
        ans.add(root.val);
    }
}