<p>Given the <code>head</code> of a singly linked list, return <code>true</code><em> if it is a </em><span data-keyword="palindrome-sequence"><em>palindrome</em></span><em> or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/pal1linked-list.jpg" style="width: 422px; height: 62px;" />
<pre>
<strong>Input:</strong> head = [1,2,2,1]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/pal2linked-list.jpg" style="width: 182px; height: 62px;" />
<pre>
<strong>Input:</strong> head = [1,2]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you do it in <code>O(n)</code> time and <code>O(1)</code> space?
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseUsingRecursion(ListNode* prev,ListNode* curr){
        if(curr==NULL){
            return prev;
        }
        ListNode* nextNode=curr->next;
        curr->next=prev;
        prev=curr;
        curr=nextNode; 

        return reverseUsingRecursion(prev,curr);
   }

   ListNode* middleNode(ListNode* head){
      ListNode* slow=head;
      ListNode* fast=head;
      while(fast->next!=NULL){
        fast=fast->next;
        if(fast->next!=NULL){
            fast=fast->next;
            slow=slow->next;
        }
      }
      return slow;
   }

   bool compare(ListNode* head1, ListNode* head2){
     while(head2!=NULL){
        if(head1->val==head2->val){
            head1=head1->next;
            head2=head2->next;
        }
        else{
            return false;
        }
     }
     return true;
   }
    bool isPalindrome(ListNode* head) {
       ListNode* mid= middleNode(head);
       ListNode* head2=mid->next;
       mid->next=NULL;
       ListNode* prev=NULL;
       ListNode* curr=head2;
       head2=reverseUsingRecursion(prev,curr);
       bool ans=compare(head,head2);
       return ans;
    }

};
50235