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# 从尾到头打印链表
 [](https://imhuay.gitbook.io/studies/algorithms/problems/2021/11/pages/R5NyOzkn3qAZy7wCx1pS#简单) [](https://imhuay.gitbook.io/studies/algorithms/problems/2021/11/pages/R5NyOzkn3qAZy7wCx1pS#剑指offer) [](https://imhuay.gitbook.io/studies/algorithms/problems/2021/11/pages/R5NyOzkn3qAZy7wCx1pS#链表) [](https://imhuay.gitbook.io/studies/algorithms/problems/2021/11/pages/R5NyOzkn3qAZy7wCx1pS#栈队列) [](https://imhuay.gitbook.io/studies/algorithms/problems/2021/11/pages/R5NyOzkn3qAZy7wCx1pS#深度优先搜索) [](https://imhuay.gitbook.io/studies/algorithms/problems/2021/11/pages/R5NyOzkn3qAZy7wCx1pS#递归)
**问题简述**
```
从尾到头打印链表（用数组返回）
```
详细描述
```
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。
```
**思路**
* 法1）利用栈，顺序入栈，然后依次出栈即可
* 法2）利用深度优先遍历思想（二叉树的先序遍历）
Python：栈
```python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
stack = []
while head:
stack.append(head.val)
head = head.next
# ret = []
# for _ in range(len(stack)): # 相当于逆序遍历
# ret.append(stack.pop())
# return ret
return stack[::-1] # 与以上代码等价
```
Python：DFS、递归
```python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
if head is None:
return []
ret = self.reversePrint(head.next)
ret.append(head.val)
return ret
```