Problem

Given the head of a doubly linked list, delete the last node and return the head of the resulting list. If the list is empty return null; if it has one node the result is empty.

Input / Output

• Input: head of a DLL.
• Output: head after deleting the tail.

Examples

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

Example 2 Input: 9 Output: empty list

Example 3 Input: 4 <-> 8 Output: 4

Constraints

• 0 <= number of nodes <= 10^5

Approach

1. If head is null, return null.
2. If head.next is null, the only node is the tail — return null.
3. Otherwise walk to the last node tail (follow next until next is null).
4. Let prev = tail.prev. Set prev.next = null and tail.prev = null.
5. Return head.

If the structure also maintains a tail pointer (common in production implementations), step 3 becomes O(1) and the whole deletion is constant time — worth mentioning as the reason doubly linked lists make good deques.

Complexity

• Time: O(n) to find the tail (O(1) with a stored tail pointer).
• Space: O(1)

Reference solution (Python)

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

Test cases

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