思路

使用递归实现

出口 ，遇到null

每一层要做：把下层放进去，把本层放下去

代码

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/class Solution {public ListNode result=new ListNode(0);public ListNode resultCur=result;public ListNode trainningPlan(ListNode head) {if (head == null) {return null;}ListNode digui = digui(head, head);return digui;}public ListNode digui(ListNode realHead,ListNode head){if (head == null) {return result;}//将后面的放入digui(realHead,head.next);//将这个本身放入resultCur.val = head.val;if (head != realHead) {//然后resultCur向后走ListNode listNode = new ListNode();resultCur.next = listNode;}resultCur = resultCur.next;return result;}}

记录

总结