590. N-ary Tree Postorder Traversal

590. N-ary Tree Postorder Traversal (Easy)

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

For example, given a 3-ary tree:

 

 

Return its postorder traversal as: [5,6,3,2,4,1].

 

Note:

Recursive solution is trivial, could you do it iteratively?

Related Topics:
Tree

Similar Questions:

• Binary Tree Postorder Traversal (Hard)
• N-ary Tree Level Order Traversal (Easy)
• N-ary Tree Preorder Traversal (Easy)

Solution 1. Recursive

// OJ: https://leetcode.com/problems/n-ary-tree-postorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(logN)
class Solution {
private:
    vector<int> ans;
    void rec(Node *root) {
        if (!root) return;
        for (auto ch : root->children) rec(ch);
        ans.push_back(root->val);
    }
public:
    vector<int> postorder(Node* root) {
        rec(root);
        return ans;
    }
};

Solution 2. Iterative

Preorder traverse the mirrored tree and reverse the result!

// OJ: https://leetcode.com/problems/n-ary-tree-postorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(logN)
class Solution {
public:
    vector<int> postorder(Node* root) {
        if (!root) return {};
        stack<Node*> s;
        s.push(root);
        vector<int> ans;
        while (s.size()) {
            root = s.top();
            s.pop();
            ans.push_back(root->val);
            for (auto c : root->children) s.push(c);
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};