逆转链表

每次我都不想接触链表和树的操作

链表结构
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

这次要求逆转链表结构(javascript)

核心思想是通过先存储链表当前节点的next数据  --------- let tt = tem.next;

使当前的节点的next指向我们设置的新链表（开始为null）---------------tem.next = newhead;

更新新链表 ---------------------newhead = tem;

更新旧链表准备下一次循环 ----------------------tem = tt;

 /**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function(head) {
        if(head === null){
            return null;
        }
        let newhead = null;
        let tem = head;
        while (tem != null){
            let tt = tem.next;
            tem.next = newhead;
            newhead = tem;
            tem = tt;
        }
        return newhead;
};