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解法一：遍历每个节点，将val存入一个list中，然后问题变成了如何反转list

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
 
class Solution:
    # 返回从尾部到头部的列表值序列，例如[1,2,3]
    def printListFromTailToHead(self, listNode):
        # write code here
        curNode = listNode
        res = []
        while curNode != None:
            res.append(curNode.val)
            curNode = curNode.next
        ret = []
        for i in range(len(res)-1,-1,-1):
            ret.append(res[i])
        return ret

解法二：反转链表
这个问题一定要掌握，同学在面腾讯时候有问到，其实就是要有三个指针，prev,cur,temp

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
 
class Solution:
    # 返回从尾部到头部的列表值序列，例如[1,2,3]
    def printListFromTailToHead(self, listNode):
        # write code here
        curNode = listNode
        res = []
        prev = None
        while curNode != None:
            #res.append(curNode.val)
            temp = curNode.next
            curNode.next = prev
            prev = curNode
            curNode = temp
        while prev != None:
            res.append(prev.val)
            prev = prev.next
        return res

/**
*  struct ListNode {
*        int val;
*        struct ListNode *next;
*        ListNode(int x) :
*              val(x), next(NULL) {
*        }
*  };
*/
class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        vector<int> res;
        ListNode* p = helper(head);
        while(p){
            res.push_back(p->val);
            p = p->next;
        }
        return res;
    }
private:
    ListNode* helper(ListNode * head){
        ListNode* prev = NULL;
        ListNode* cur = head;
        while(cur){
            ListNode* tmp = cur->next;
            cur->next = prev;
            prev = cur;
            cur = tmp;
        }
        return prev;
    }
};

解法三：其实第一种解法相当于是将数据存入一个栈，那么既然存入了栈，我们就可以用递归的方法来解决（这是书里的思路，重点：栈——》递归）

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # 返回从尾部到头部的列表值序列，例如[1,2,3]
    def printListFromTailToHead(self, listNode):
        # write code here
        Head = listNode
        res = []
        def recursively(pHead):
            if pHead != None:
                if pHead.next != None:
                    recursively(pHead.next)
                res.append(pHead.val)
        recursively(Head)
        return res

c++解法：

/**
*  struct ListNode {
*        int val;
*        struct ListNode *next;
*        ListNode(int x) :
*              val(x), next(NULL) {
*        }
*  };
*/
class Solution {
public:
    vector<int> res;
    vector<int> printListFromTailToHead(ListNode* head) {
        helper(head);
        return res;
    }
private:
    void helper(ListNode *head){
        if(head == NULL)
            return;
        helper(head->next);
        res.push_back(head->val);
    }
};