Traversal and Printing a Linked List in DSA C++ - 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 the items in a linked list and print them.

The question is: how does the time needed grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct Node {
    int data;
    Node* next;
};

void printList(Node* head) {
    Node* current = head;
    while (current != nullptr) {
        std::cout << current->data << " -> ";
        current = current->next;
    }
    std::cout << "null" << std::endl;
}
    

This code goes through each node in the linked list from the start and prints its value until it reaches the end.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once for each node in the list, from first to last.

How Execution Grows With Input

As the list gets longer, the time to print grows directly with the number of nodes.

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

Pattern observation: The time grows in a straight line as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to print grows directly in proportion to the number of nodes in the list.

Common Mistake

[X] Wrong: "Printing a linked list is faster than printing an array because it uses pointers."

[OK] Correct: Both linked lists and arrays require visiting each element once to print, so they take similar time proportional to their size.

Interview Connect

Understanding how to traverse and print a linked list efficiently is a basic but important skill that shows you know how to work with pointers and data structures.

Self-Check

"What if we changed the linked list to a doubly linked list? How would the time complexity of printing change?"

Practice

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

easy

A. To delete all nodes at once

B. To reverse the linked list automatically

C. To create a new linked list

D. To visit each node and access its data

Solution

1. Step 1: Understand traversal meaning

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

2. Step 2: Purpose of traversal

   The main goal is to access or process the data stored in each node during the visit.

3. Final Answer:

   To visit each node and access its data -> 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 new lists
• Assuming traversal reverses the list

2. Which of the following is the correct syntax to start traversing a singly linked list in C++ if head is the pointer to the first node?

easy

A. Node* current = head;

B. Node current = head;

C. Node* current = &head;

D. Node current = *head;

Solution

1. Step 1: Understand pointer usage

   In C++, linked list nodes are accessed via pointers. head is a pointer to the first node.

2. Step 2: Correct pointer assignment

   To traverse, we create a pointer current and assign it to head directly: Node* current = head;

3. Final Answer:

   Node* current = head; -> Option A

4. Quick Check:

   Use pointer to start traversal [OK]

Hint: Use pointer assignment to start traversal [OK]

Common Mistakes:

• Using object instead of pointer
• Taking address of head incorrectly
• Dereferencing head wrongly

3. What will be the output of the following C++ code snippet that prints a linked list?

struct Node {
  int data;
  Node* next;
};

int main() {
  Node n3 = {3, nullptr};
  Node n2 = {2, &n3};
  Node n1 = {1, &n2};
  Node* head = &n1;

  Node* current = head;
  while (current != nullptr) {
    std::cout << current->data << " -> ";
    current = current->next;
  }
  std::cout << "null" << std::endl;
  return 0;
}

medium

A. null -> 1 -> 2 -> 3

B. 3 -> 2 -> 1 -> null

C. 1 -> 2 -> 3 -> null

D. 1 2 3 null

Solution

1. Step 1: Trace node connections

   Node n1 points to n2, n2 points to n3, n3 points to nullptr (end).

2. Step 2: Follow the while loop output

   The loop prints data of current node then moves to next. So prints 1, then 2, then 3, then stops.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option C

4. Quick Check:

   Print nodes in order from head [OK]

Hint: Follow next pointers from head until nullptr [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing '->' arrows in output
• Not printing 'null' at the end

4. Identify the error in the following code snippet that attempts to print a linked list:

struct Node {
  int data;
  Node* next;
};

void printList(Node* head) {
  Node* current = head;
  while (current->next != nullptr) {
    std::cout << current->data << " -> ";
    current = current->next;
  }
  std::cout << "null" << std::endl;
}

medium

A. The loop condition should be current != nullptr, not current->next != nullptr

B. The print statement should be after the loop

C. The function should return an int

D. The pointer current should be declared as Node current, not Node* current

Solution

1. Step 1: Analyze loop condition

   The loop stops when current->next is nullptr, so it skips printing the last node's data.

2. Step 2: Correct loop condition

   It should check current != nullptr to print all nodes including the last one.

3. Final Answer:

   The loop condition should be current != nullptr, not current->next != nullptr -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Stopping loop too early missing last node
• Wrong pointer type declaration
• Misplacing print statement

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 or using recursion?

hard

A. Traverse normally and print nodes as you go

B. Use a stack to store nodes during traversal, then pop and print

C. Reverse the list first, then print

D. Use two pointers to print backwards

Solution

1. Step 1: Understand constraints

   We cannot change the list or use recursion, so reversing pointers or recursive calls are not allowed.

2. Step 2: Use auxiliary data structure

   Use a stack to push each node's data during traversal, then pop from stack to print in reverse order.

3. Final Answer:

   Use a stack to store nodes during traversal, then pop and print -> Option B

4. Quick Check:

   Stack reverses order without modifying list [OK]

Hint: Use stack to reverse print without changing list [OK]

Common Mistakes:

• Trying to reverse list pointers directly
• Using recursion despite restriction
• Assuming two pointers can print backwards