1 public ListNode ReverseList(ListNode head) {
2     if (head == null || head.next == null)
3         return head;
4     ListNode next = head.next;
5     head.next = null;
6     ListNode newHead = ReverseList(next);
7     next.next = head;
8     return newHead;
9 }

 1 class Solution {
 2     public ListNode reverseList(ListNode head) {
 3         //申请节点，pre和 cur，pre指向null
 4         ListNode pre = null;
 5         ListNode cur = head;
 6         ListNode tmp = null;
 7         while(cur!=null) {
 8             //记录当前节点的下一个节点
 9             tmp = cur.next;
10             //然后将当前节点指向pre
11             cur.next = pre;
12             //pre和cur节点都前进一位
13             pre = cur;
14             cur = tmp;
15         }
16         return pre;
17     }
18 }