<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* 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;
    }

    ListNode* reverseUsingRecursion(ListNode* prev, ListNode* curr) {
        //base case
        if(curr == NULL) {
            return prev;
        }
        //1 case hum solve krenge
        ListNode* nextNode = curr->next;
        curr->next = prev;
        prev = curr;
        curr = nextNode;

        //baaaki kon sambhalega - recursion
        return reverseUsingRecursion(prev, curr);
    }

    bool compareList(ListNode* head1, ListNode* head2) {
        while(head1 != NULL && head2 != NULL) {
            if(head1->val != head2->val) {
                return false;
            }
            else {
                head1 = head1->next;
                head2 = head2->next;
            }
        }
        return true;
    }

    bool isPalindrome(ListNode* head) {
        //break into 2 havles
        ListNode* midNode = middleNode(head);
        ListNode* head2 = midNode -> next;
        midNode -> next = NULL;

        //reverse second half
        ListNode* prev = NULL;
        ListNode* curr = head2;
        head2 = reverseUsingRecursion(prev, curr);
        
        //compare both linked list
        bool ans = compareList(head, head2);
        return ans;
    }
};
51825