LeetCode #590. N-ary Tree Postorder Traversal

Level: Easy

Given the root of an n-ary tree, return the postorder traversal of its nodes’ values.

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

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [5,6,3,2,4,1]

Example 2:

Input: 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]
Output: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• 0 <= Node.val <= 104
• The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?

My Solution:

//postorder left->right->root
class Solution {
public:
    vector<int> postorder(Node* root) {
        if(!root) return {};
        vector<int> ans;
        stack<Node*> st;
        st.push(root);
        while(!st.empty()){
            Node* node = st.top();
            st.pop();
            for(Node* n : node->children) st.push(n);
            ans.push_back(node->val);
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};