Traversal and Printing a Linked List in DSA Java - Time & Space Complexity

Time Complexity: Traversal and Printing a Linked List

O(n)

Understanding Time Complexity

We want to understand how long it takes to go through all items in a linked list and print them.

How does the time needed change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node {
  int data;
  Node next;
}

void printList(Node head) {
  Node current = head;
  while (current != null) {
    System.out.println(current.data);
    current = current.next;
  }
}
    

This code goes through each node in the linked list from start to end and prints its data.

Identify Repeating Operations

• Primary operation: Visiting each node once to print its data.
• How many times: Exactly once per node, so as many times as there are nodes in the list.

How Execution Grows With Input

As the list grows, the number of print steps grows the same way.

Input Size (n)	Approx. Operations
10	10 visits and prints
100	100 visits and prints
1000	1000 visits and prints

Pattern observation: The work grows directly with the number of nodes. Double the nodes, double the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to print the list grows in a straight line with the number of nodes.

Common Mistake

[X] Wrong: "Since we only print, the time is constant no matter the list size."

[OK] Correct: Each node must be visited and printed, so more nodes mean more steps, not a fixed time.

Interview Connect

Knowing how to analyze simple list traversal helps you understand more complex list operations and shows you can reason about code efficiency clearly.

Self-Check

"What if we printed only every other node? How would the time complexity change?"

Practice

1. What is the main purpose of traversing a linked list in Java?

easy

A. To sort the linked list nodes automatically

B. To delete the entire linked list at once

C. To create a new linked list from scratch

D. To visit each node and perform an action like printing its value

Solution

1. Step 1: Understand traversal meaning

   Traversal means visiting each node one by one in the linked list.

2. Step 2: Identify traversal purpose

   Commonly, traversal is used to perform actions like printing or searching nodes.

3. Final Answer:

   To visit each node and perform an action like printing its value -> Option D

4. Quick Check:

   Traversal = Visit nodes [OK]

Hint: Traversal means visiting nodes one by one [OK]

Common Mistakes:

• Confusing traversal with deletion
• Thinking traversal creates a new list
• Assuming traversal sorts the list

2. Which of the following is the correct syntax to start traversing a singly linked list named head in Java?

easy

A. Node current = head.next;

B. Node current = null;

C. Node current = head;

D. Node current = head.prev;

Solution

1. Step 1: Identify starting point of traversal

   Traversal starts from the first node, which is referenced by head.

2. Step 2: Choose correct initialization

   We assign current to head to begin traversal at the first node.

3. Final Answer:

   Node current = head; -> Option C

4. Quick Check:

   Start traversal at head node [OK]

Hint: Start traversal from head node reference [OK]

Common Mistakes:

• Starting from head.next skips first node
• Initializing current to null causes no traversal
• Using head.prev in singly linked list causes error

3. What will be the output of the following Java code snippet?

class Node {
  int data;
  Node next;
  Node(int d) { data = d; next = null; }
}

public class Main {
  public static void main(String[] args) {
    Node head = new Node(5);
    head.next = new Node(10);
    head.next.next = new Node(15);

    Node current = head;
    while (current != null) {
      System.out.print(current.data + " -> ");
      current = current.next;
    }
    System.out.print("null");
  }
}

medium

A. 5 -> 10 -> 15 -> null

B. null -> 5 -> 10 -> 15

C. 5 10 15 null

D. 5 -> 10 -> null

Solution

1. Step 1: Trace node creation and linking

   Nodes created with data 5, 10, 15 linked as 5 -> 10 -> 15 -> null.

2. Step 2: Follow while loop printing

   Loop prints each node's data followed by ' -> ', then ends with 'null'.

3. Final Answer:

   5 -> 10 -> 15 -> null -> Option A

4. Quick Check:

   Print nodes in order with arrows [OK]

Hint: Print each node data with arrow until null [OK]

Common Mistakes:

• Printing null at start instead of end
• Missing arrows between nodes
• Stopping before last node

4. Identify the error in the following code snippet for printing a linked list and select the correct fix.

Node current = head;
while (current.next != null) {
  System.out.print(current.data + " -> ");
  current = current.next;
}
System.out.print("null");

medium

A. Remove System.out.print("null") line

B. Change condition to current != null to print last node

C. Initialize current to head.next to skip first node

D. Add current = current.next inside the loop

Solution

1. Step 1: Analyze loop condition

   The loop stops when current.next is null, so last node is not printed.

2. Step 2: Fix condition to include last node

   Changing condition to current != null ensures last node is printed.

3. Final Answer:

   Change condition to current != null to print last node -> Option B

4. Quick Check:

   Loop until current is null to print all nodes [OK]

Hint: Loop until current is null to include last node [OK]

Common Mistakes:

• Using current.next != null misses last node
• Skipping first node by wrong initialization
• Removing null print causes unclear output

5. Given a singly linked list, how would you modify the traversal code to print the list in reverse order without changing the list structure?

hard

A. Use recursion to traverse to the end, then print nodes on return

B. Reverse the list first, then traverse normally

C. Use a loop to print nodes from tail to head directly

D. Print nodes while moving forward, then print again backward

Solution

1. Step 1: Understand limitation of singly linked list

   Singly linked list nodes only point forward, so backward traversal is not direct.

2. Step 2: Use recursion to print in reverse

   Recursion goes to end first, then prints nodes on returning back, achieving reverse print without changing list.

3. Final Answer:

   Use recursion to traverse to the end, then print nodes on return -> Option A

4. Quick Check:

   Recursion prints reverse without modifying list [OK]

Hint: Recursion prints reverse by returning from end [OK]

Common Mistakes:

• Trying to loop backward in singly linked list
• Reversing list changes structure, not allowed here
• Printing twice forward and backward duplicates output