来源：力扣（LeetCode） 链接：https://leetcode.cn/problems/reverse-linked-list 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。
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解答

迭代解决

  1. /**
  2.  * Definition for singly-linked list.
  3.  * function ListNode(val, next) {
  4.  *     this.val = (val===undefined ? 0 : val)
  5.  *     this.next = (next===undefined ? null : next)
  6.  * }
  7.  */
  8. /**
  9.  * @param {ListNode} head
  10.  * @return {ListNode}
  11. */
  12. var reverseList = function(head) {
  13.     let prevNode = null;
  14.     while (head) {
  15.         const next = head.next;
  16.         head.next = prevNode;
  17.         prevNode = head;
  18.         head = next;
  19.     }
  20.     return prevNode;
  21. };

递归解决

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
*/
var reverseList = function(head) {
    let firstNode = null;
    function reverseNode (node, prevNode) {
        if (!node) return;
        const next = node.next;
        if (next === null) {
            firstNode = node;
        }
        node.next = prevNode;
        reverseNode(next, node);
    }
    reverseNode(head, null);
    return firstNode;
};