<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 {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

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

class Solution {
public:
void f(Node* root,vector<int> &ans){
    if(root==NULL) return;
    ans.push_back(root->val);
    for(auto it:root->children){
        f(it,ans);
    }
    return ;
}
    vector<int> preorder(Node* root) {
        if(root==NULL) return {};
        stack<Node*> st;
        stack<int> num;
        st.push(root);
        while(!st.empty()){
            Node* temp=st.top();
            st.pop();
            stack<Node*> que;
            num.push(temp->val);
            for(auto it:temp->children){
                que.push(it);
            }
            while(!que.empty()){
                st.push(que.top());
                que.pop();
            }
        }
        vector<int> ans;
        while(!num.empty()){
            ans.push_back(num.top());
            num.pop();
        }
        reverse(ans.begin(),ans.end());
        return ans;
    }
};
25600