Same solution as N-ary Tree PreOrder Traversal and just reversing the result
Faster than 91% \n\n\nclass Solution {\npublic:\n    vector<int> postorder(Node* root) {\n        vector<int> out;\n        \n        if(!root)\n            ret
**Faster than 91% **\n\n```\nclass Solution {\npublic:\n    vector<int> postorder(Node* root) {\n        vector<int> out;\n        \n        if(!root)\n            return out;\n        \n        stack<Node*> s;\n        s.push(root);\n        \n        while(!s.empty()){\n            Node* node = s.top(); s.pop();\n            out.push_back(node->val);\n            \n            for(int i = 0; i < node->children.size(); i++){\n                s.push(node->children[i]);\n            }\n        }\n        reverse(out.begin(), out.end());\n        return out;\n    }\n};\n```
['C', 'Iterator', 'C++']