方案一

利用链表的后续遍历，使用函数调用栈作为后序遍历栈，来判断是否回文

复制
1var isPalindrome = function(head) {
2    let left = head;
3    function traverse(right) {
4        if (right == null) return true;
5        let res = traverse(right.next);
6        res = res && (right.val === left.val);
7        left = left.next;
8        return res;
9    }
10    return traverse(head);
11};

方案二

通过快、慢指针找链表中点，然后反转链表，比较两个链表两侧是否相等，来判断是否是回文链表

复制
1var isPalindrome = function(head) {
2    // 反转 slower 链表
3    let right = reverse(findCenter(head));
4    let left = head;
5    // 开始比较
6    while (right != null) {
7        if (left.val !== right.val) {
8            return false;
9        }
10        left = left.next;
11        right = right.next;
12    }
13    return true;
14}
15function findCenter(head) {
16    let slower = head, faster = head;
17    while (faster && faster.next != null) {
18        slower = slower.next;
19        faster = faster.next.next;
20    }
21    // 如果 faster 不等于 null，说明是奇数个，slower 再移动一格
22    if (faster != null) {
23        slower = slower.next;
24    }
25    return slower;
26}
27function reverse(head) {
28    let prev = null, cur = head, nxt = head;
29    while (cur != null) {
30        nxt = cur.next;
31        cur.next = prev;
32        prev = cur;
33        cur = nxt;
34    }
35    return prev;
36}
37