Problem URL: Palindrome linked list

We will iterate over the whole list, and add the result to an array. Then we check whether the array is palindrome or not and return that value.

# 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:
        values = []
        while head:
            values.append(head.val)
            head = head.next

        l, r = 0, len(values)-1
        while l < r:
            if values[l] != values[r]:
                return False
            l += 1
            r -= 1
        return True

Time Complexity: O(n)
Space Complexity: O(n)