Which describes the classic approach to check if a singly linked list is a palindrome using a stack?

• A. Use two pointers with no extra storage
• B. Convert the list to a string and compare
• C. Push the first half values onto a stack while finding the middle, then compare with the second half; O(n) time, O(n) space
• D. Reverse the list and compare with the original

Answer: (C)

Using a stack to check a palindrome works because a stack gives you access to elements in reverse order. The idea is to read the list up to the middle, pushing each value onto a stack. This stores the first half in the order that will be reversed when you pop. Then you continue from the middle to the end, popping values from the stack and comparing to the current node. If all pairs match, the list reads the same forward and backward.

Handling the middle is important: in an even-length list you’ll have a clear second half to compare with the reversed first half; in an odd-length list you skip the middle node, since it doesn’t affect palindromicity.

This approach runs in linear time, because you traverse the list a couple of passes in total, and it uses linear extra space because the stack holds about half of the nodes. It’s a classic, straightforward way to use a stack to expose the first-half values in reverse order for direct comparison with the second half, without altering the list structure.