N-ary Tree Postorder Traversal

Description

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?

Code

HRL 的 reverse 就是 LRH，即是 post order traversal。要找 HRL，就和 N-ary Tree Preorder Traversal 類似了。

/*
// 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) return {};
        stack<Node*> st;
        vector<int> answer;
        st.push(root);
        while(!st.empty()) {
            auto node = st.top();
            st.pop();
            answer.emplace_back(node->val);
            for(auto child: node->children) {
                st.push(child);
            }
        }
 
        reverse(answer.begin(), answer.end());
        return answer;
    }
};

Source

• N-ary Tree Postorder Traversal - LeetCode