反转链表-python

leetCode第206题 反转链表。
链接：https://leetcode-cn.com/problems/reverse-linked-list
给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

示例1：

示例2:

示例 3：

输入：head = []
输出：[]

提示：

链表中节点的数目范围是 [0, 5000]
-5000 <= Node.val <= 5000

提示：

链表中节点的数目范围是 [0, 5000]
-5000 <= Node.val <= 5000

进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？

## python3
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head == None:
            return None
        pre = None
        cur = head
        nex = head.next
        while cur != None:
            cur.next = pre
            cur = nex
            nex = nex.next
        return pre

## python3
class Solution1:
    tail = ListNode()
    def reverseList(self, head: ListNode) -> ListNode:
        if head == None:
            return head
        head = self.digui(head)
        return self.tail
    def digui(self, head: ListNode) -> ListNode:
        if head.next == None:
            self.tail = head
            return head
        else:
            temp = self.digui(head.next)
            temp.next = head
            head.next = None
            return head