<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://assets.leetcode.com/uploads/2018/10/12/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://assets.leetcode.com/uploads/2019/11/08/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>

/*
// 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:
    vector<int> postorder(Node* root) 
    {
        if (root == nullptr)
        {
            return {};
        }
        vector<int> res;
        backtrack(root, res);
        return res;
    }

    void backtrack(Node* curr, vector<int> &res)
    {
        deque<Node*> notExplored;
        if (curr == nullptr)
        {
            return;
        }
        else if (curr->children.size() != 0)
        {
            // This will get all the children
            for (auto it: curr->children)
            {
                notExplored.push_back(it);
            }
        }

        while (notExplored.size() != 0)
        {
            Node* explore = notExplored.front();
            notExplored.pop_front();
            backtrack(explore, res);
        }
        res.push_back(curr->val);
    }
};

    
// Given the root of an n-ary tree, return the postorder traversal of its nodes' values
// What is post order
    // Visit left
    // Visit right
    // Record
// Temp solution's thoughts
    // In the current stack, get all the current children
    // Create a new stack by entering left
    // Our idea is each node will have its own storage to keep track of its children
    // When we want to explore a children, we take the node out of the current node's storage
    // When a node does not have anything left in its storage, we would want to add them to the vector
// Solution
    // Function should take in the current node, the solution array
    // Base case
        // If the node is null, just return an empty array or something
        // If the node does not have children, we would want to add that to our current array solution
        // If the node has children, we would want to add them to the list of stuff to explore
    // Recursive case
        // If the storage of the current node has children, remove that child and explore that child
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