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);}
};