# Palindrome Linked List

Given the head of a singly linked list, return true* if it is a palindrome or *false* otherwise*.

```
class Solution:
def isPalindrome(self, head: ListNode) -> bool:
slow, fast, prev = head, head, None
while fast and fast.next:
slow, fast = slow.next, fast.next.next
prev, slow, prev.next = slow, slow.next, None
while slow:
slow.next, prev, slow = prev, slow, slow.next
fast, slow = head, prev
while slow:
if fast.val != slow.val: return False
fast, slow = fast.next, slow.next
return True
```

## Related Problems

Given the head of a singly linked list, reverse the list, and return *the reversed list*.

Given the head of a linked list, remove the nth node from the end of the list and return its head.

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.

Return *the head of the merged linked list*.

Given the root of a binary tree, flatten the tree into a "linked list":

The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.

The "linked list" should be in the same order as a pre-order traversal of the binary tree.