<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 reverseList(self,head):
        if head == None or head.next == None:
            return head

        newHead = self.reverseList(head.next)
        if head.next != None:
            head.next.next = head

        return newHead


    """
    def reverseList(self, head):
        return self._reverse(head)

    def _reverse(self, node, prev=None):
        if not node:
            return prev
        n = node.next
        node.next = prev
        return self._reverse(n, node)
    """
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        slow = head
        fast = head
        prev = None
        while fast and fast.next != None:
            prev = slow
            slow = slow.next
            fast = fast.next.next

        node_to_reverse = slow
        reversed_head = self.reverseList(node_to_reverse)

        if fast == None:
            prev.next = None
        else:
            slow.next = None

        while reversed_head != None and head != None:
            if reversed_head.val != head.val:
                return False

            reversed_head = reversed_head.next
            head = head.next
        
        return True

        
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