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234. Palindrome Linked List

Easy

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Given a singly linked list, determine if it is a palindrome.

Example 1:

Input: 1->2
Output: false

Example 2:

Input: 1->2->2->1
Output: true

Follow up:
Could you do it in O(n) time and O(1) space?

class Solution {
public:
    ListNode* GetMid(ListNode* head){
        ListNode* Fast = head;
        ListNode* Slow = head;
        while(Fast->next != NULL && Fast->next->next != NULL){
            Slow = Slow->next;
            Fast = Fast->next->next;
        }
        return Slow;
    }
    ListNode* Reverse(ListNode* mid){
        ListNode* PreNode = NULL;
        ListNode* CurrentNode = mid->next;
        ListNode* ReverseHead = NULL;
        while(CurrentNode != NULL){
            ListNode* NextNode = CurrentNode->next;
            if(NextNode == NULL){
                ReverseHead = CurrentNode;
            }
            CurrentNode->next = PreNode;
            PreNode = CurrentNode;
            CurrentNode = NextNode;
        }
        return ReverseHead;
    }
    bool isPalindrome(ListNode* head) {
        if(head == NULL ||head->next == NULL){
            return true;
        }
        ListNode* MiddleNode = GetMid(head);
        ListNode* ReverseHead = Reverse(MiddleNode);
        while(head != NULL && ReverseHead != NULL){
            if(head->val != ReverseHead->val){
                return false;
            }
            head = head->next;
            ReverseHead = ReverseHead->next;
        }
        return true;
    }
};