Problem

Given the head of a singly linked list, delete the last node and return the head. If the list is empty or has one node, return null.

Input / Output

• Input: head of a singly linked list.
• Output: head after deleting the tail.

Examples

Example 1 Input: 1 -> 2 -> 3 Output: 1 -> 2

Example 2 Input: 9 Output: empty

Example 3 Input: 4 -> 8 Output: 4

Constraints

• 0 <= number of nodes <= 10^5

Approach

Because nodes only point forward, deleting the tail requires the second-to-last node:

1. If head is null or head.next is null, return null.
2. Walk cur until cur.next.next is null — cur is now the second-to-last node.
3. Set cur.next = null.

Unlike a doubly linked list (where a stored tail pointer makes this O(1)), a singly linked list needs the full traversal even if a tail pointer exists, because the new tail's predecessor is unknown. Interviewers use this contrast to ask when would you choose a doubly linked list?

Complexity

• Time: O(n)
• Space: O(1)

Reference solution (Python)

def delete_tail(head):
    if head is None or head.next is None:
        return None
    cur = head
    while cur.next.next:
        cur = cur.next
    cur.next = None
    return head

Test cases

Input	Expected
1->2->3	1->2
9	None
4->8	4
empty	None