590. N-ary Tree Postorder Traversal*

590. N-ary Tree Postorder Traversal*

https://leetcode.com/problems/n-ary-tree-postorder-traversal/

题目描述

Given 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).

Follow up:

Recursive solution is trivial, could you do it iteratively?

Example 1:

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

Example 2:

```bash 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 height of the n-ary tree is less than or equal to 1000
• The total number of nodes is between [0, 10^4]

C++ 实现 1

递归的方式.

/*
// 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 {};
        vector<int> res;
        for (auto &c : root->children) {
            auto tmp = postorder(c);
            std::copy(tmp.begin(), tmp.end(), back_inserter(res));
        }
        res.push_back(root->val);
        return res;
    }
};

C++ 实现 2

迭代的方式. 迭代的方式可以仔细研究一下, children 按顺序入栈, 最后 reverse res.

class Solution {
public:
    vector<int> postorder(Node* root) {
        if (!root) return {};
        vector<int> res;
        stack<Node*> st;
        st.push(root);
        while (!st.empty()) {
            auto r = st.top();
            st.pop();
            res.push_back(r->val);
            for (auto &c : r->children) st.push(c);
        }
        std::reverse(res.begin(), res.end());
        return res;
    }
};