Leetcode刷题笔记题解（C++）：590. N 叉树的后序遍历

思路：类似于二叉树的排序，这里需要将子树进行依次递归遍历，前序遍历也与之类似

/*
// 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:
    //用于保存结果
    vector<int> res;
    vector<int> postorder(Node* root) {
        helper(root);
        return res;
    }
    //递归函数
    void helper(Node* root){
        //如果节点为空
        if(root == nullptr) return;
        //获取当前节点的子节点
        vector<Node*> children = root->children;
        //依次递归调用叶子节点
        for(int i = 0;i<children.size();i++){
            helper(children[i]);
        }
        res.push_back(root->val);
    }
};