用递归进行链表逆置

 1 node* reverse(node* head , node* frOnt= NULL)//head为头指针，调用此函数时只需传入第一个参数。返回值是逆置后链表的头指针
 2 {
 3     if(head == NULL)
 4     {
 5         return head ;
 6     }
 7     if(head->next != NULL)
 8     {
 9         node* temp = reverse(head->next , head) ;
10         head->next = front ;
11         return temp ;
12     }
13     else
14     {
15         head->next = front ;
16         return head ;
17     }
18 }