Problem

Given the head of a singly linked list and an integer val, append a new node with value val at the end of the list and return the head. If the list is empty, the new node becomes the head.

Input / Output

• Input: head (possibly null), integer val.
• Output: head of the updated list.

Examples

Example 1 Input: 1 -> 2 -> 3, val = 4 Output: 1 -> 2 -> 3 -> 4

Example 2 Input: empty, val = 9 Output: 9

Constraints

• 0 <= number of nodes <= 10^5

Approach

1. If head is null, return a new node.
2. Walk cur until cur.next is null (the last node).
3. Set cur.next = ListNode(val) and return head.

Appending costs O(n) because a singly linked list has no direct reference to its last node. In practice, list implementations keep a tail pointer so that append becomes O(1) — a good point to raise. With a tail pointer: tail.next = node; tail = node.

Follow-up: append k values in one pass by keeping cur at the tail after the first append rather than re-walking each time.

Complexity

• Time: O(n) (O(1) with a tail pointer)
• Space: O(1)

Reference solution (Python)

def insert_tail(head, val):
    node = ListNode(val)
    if head is None:
        return node
    cur = head
    while cur.next:
        cur = cur.next
    cur.next = node
    return head

Test cases

Input	Expected
1->2->3, val=4	[1,2,3,4]
empty, val=9	[9]
7, val=8	[7,8]
append twice to empty (1 then 2)	[1,2]