How do you traverse a singly linked list to print all node values, and what is the time complexity?

• A. Follow next from head to null, printing as you go; O(n)
• B. Access nodes by index in random order; O(1) per access
• C. Traverse from tail to head using a back-pointer; O(n)
• D. Use a separate array mirroring the list; O(1)

Answer: (A)

The essential idea here is how to move through a singly linked list to print every value and what that costs. In a singly linked list, each node has a next pointer to the following node. To print all values, start at the head, print the current node’s value, then follow the next pointer to the next node, repeating until you reach null. This straightforward forward walk visits each node exactly once, so the total work grows linearly with the number of nodes. That gives O(n) time. You only need a single pointer to keep track of your position, so the extra space is O(1).

The other approaches aren’t as fitting for a standard singly linked list. Randomly accessing by index isn’t constant-time in a singly linked list—you’d have to start at the head and step through until you reach the desired position, so printing all values isn’t effectively constant per element. Traversing from tail to head requires a back-pointer or a doubly linked structure, which a plain singly linked list doesn’t provide. Mirroring the list into another array also adds extra space and isn’t necessary for simply printing all values.