1. 题目

输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。

示例 1：
输入：head = [1,3,2]
输出：[2,3,1]限制：
0 <= 链表长度 <= 10000

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

2.1 stack解题

class Solution {
public:vector<int> reversePrint(ListNode* head) {stack<int> s;while(head){s.push(head->val);head = head->next;}vector<int> ans;while(!s.empty()){ans.push_back(s.top());s.pop();}return ans;}
};

2.2 递归

class Solution {vector<int> ans;
public:vector<int> reversePrint(ListNode* head) {dfs(head);return ans;}void dfs(ListNode* head){if(!head)return;dfs(head->next);ans.push_back(head->val);}
};

2.3 反转链表

class Solution {vector<int> ans;
public:vector<int> reversePrint(ListNode* head) {if(!head)return {};head = reverseList(head);while(head){ans.push_back(head->val);head = head->next;}return ans;}ListNode* reverseList(ListNode* head){	//反转链表，返回新表头ListNode *prev = NULL, *nt = head->next;while(head && head->next){head->next = prev;prev = head;head = nt;nt = nt->next;}head->next = prev;return head;}    
};