剑指 Offer 06. 从尾到头打印链表
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

解题思路
法一：回溯法
法二：用栈
法三：利用vector的insert特性

代码
1.回溯法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void method(vector<int>& res,ListNode* head){
        if(head==NULL) return;
        method(res,head->next);
        res.push_back(head->val);
    }
    vector<int> reversePrint(ListNode* head) {
        vector<int> res;
        method(res,head);
        return res;
    }
};

2.用栈

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> reversePrint(ListNode* head) {
        stack<int> s;
        vector<int> res;
        ListNode* pre=head;
        while(pre){
            s.push(pre->val);
            pre=pre->next;
        }
        while(!s.empty()){
            res.push_back(s.top());
            s.pop();
        }
        return res;
    }
};

3.利用vector的insert特性

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> reversePrint(ListNode* head) {
        vector<int> res;
        ListNode* pre=head;
        while(pre){
            res.insert(res.begin(),pre->val);
            pre=pre->next;
        }
        return res;
    }
};

作者：junlin-v
链接：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/solution/san-chong-fang-fa-by-junlin-v-r8ar/