题解 | #反转链表#

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode ReverseList(ListNode head) {
       //通过递归做：将链表看成特殊二叉树
       //截至条件：当为叶子节点或者是节点为空截至
       if(head==null||head.next==null){
        return  head;
       }
       //先将head。next后面的全部逆置,返回头节点
       
       ListNode newhead = ReverseList(head.next);
    //将当前节点添加到逆置后的链表后面，就是改变这两个节点的指向
       head.next.next = head;
       head.next = null;
        return newhead;

    }
}