<p>Given the <code>root</code> of an n-ary tree, return <em>the preorder 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>

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

"""
# 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 preorder(self, root: 'Node') -> List[int]:
        def helper(node):
            if node:
                yield node.val
                for child in node.children:
                    yield from helper(child)
        return list(helper(root))
18100