<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.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        def reverseLL(head):
            if not head or not head.next:
                return head
            temp=reverseLL(head.next)
            head.next.next=head
            head.next=None
            return temp

        def length(head):
            count=0
            while(head):
                head=head.next
                count+=1
            return count

        def getMiddle(head):
            slow=head
            fast=head.next
            while slow and fast and fast.next:
                slow=slow.next
                fast=fast.next.next
            return slow

        def printLL(head):
            while(head):
                print(head.val,end=" ")
                head=head.next
            print()

        n=length(head)
        left=getMiddle(head)
        right=left.next
        left.next=None
        left=reverseLL(head)

        # printLL(left)
        # printLL(right)

        if n&1:
            left=left.next

        while(left and right):
            if left.val!=right.val:
                return False
            left=left.next
            right=right.next
        
        return True if (not left and not right) else False

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