Traversal and Printing a Linked List in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main purpose of traversing a linked list?

To visit each node in the linked list one by one, usually to read or process the data stored in each node.

beginner

In a singly linked list, how do you move from one node to the next during traversal?

By following the 'next' pointer/reference stored in the current node to reach the next node.

beginner

What condition is used to stop traversing a linked list?

When the current node pointer becomes NULL, meaning there are no more nodes to visit.

beginner

Why is it important to check if the head of the linked list is NULL before traversal?

Because a NULL head means the list is empty, so there are no nodes to traverse or print.

beginner

What is the output format when printing a linked list like 1 -> 2 -> 3 -> NULL?

It shows the data in each node separated by arrows (->) pointing to the next node, ending with NULL to indicate the list end.

What does the 'next' pointer in a linked list node represent?

A The address of the next node

B The data stored in the node

C The address of the previous node

D The size of the node

The 'next' pointer stores the address of the next node in the list, allowing traversal.

When traversing a linked list, what indicates you have reached the end?

A The current node's data is zero

B The current node's next pointer is NULL

C The current node's next pointer points to head

D The current node is the first node

A NULL next pointer means there is no next node, so traversal ends.

Which of these is the correct way to start traversing a linked list?

A Start from the last node

B Start from the middle node

C Start from the head node

D Start from a NULL pointer

Traversal always starts from the head node of the linked list.

What happens if you try to traverse a linked list with a NULL head?

A You traverse all nodes

B You traverse nodes in reverse

C You traverse only the last node

D You get an error or no output because the list is empty

A NULL head means the list is empty, so traversal cannot proceed.

How do you print the data of each node during traversal?

A Print the data and move to the next node until NULL

B Print only the head node data

C Print the data of the last node only

D Print the data in reverse order

You print each node's data and move to the next node until you reach NULL.

Explain step-by-step how to traverse and print all elements of a singly linked list.

Think about visiting each node one by one until the end.

Describe what the output looks like when printing a linked list with nodes containing 5, 10, and 15.

Visualize the chain of nodes connected by arrows.

Practice

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

easy

A. To delete the entire linked list

B. To visit each node and access its data

C. To create a new linked list

D. To reverse the linked list

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

   The main goal is to access or print the data stored in each node.

3. Final Answer:

   To visit each node and access its data -> Option B

4. Quick Check:

   Traversal = Visit nodes [OK]

Hint: Traversal means visiting nodes to read data [OK]

Common Mistakes:

• Confusing traversal with deletion
• Thinking traversal creates new nodes
• Assuming traversal reverses the list

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

easy

A. Node* temp = &head;

B. struct Node temp = head;

C. struct Node* temp = head;

D. Node temp = head;

Solution

1. Step 1: Understand pointer usage

   In C, linked list nodes are accessed via pointers, so temp must be a pointer to Node.

2. Step 2: Check syntax correctness

   struct Node* temp = head; correctly declares temp as a pointer and assigns it to head.

3. Final Answer:

   struct Node* temp = head; -> Option C

4. Quick Check:

   Pointer assignment = struct Node* temp = head [OK]

Hint: Use pointer variable to traverse linked list [OK]

Common Mistakes:

• Using struct Node instead of pointer
• Taking address of head incorrectly
• Missing struct keyword if needed

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

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

int main() {
  struct Node n3 = {30, NULL};
  struct Node n2 = {20, &n3};
  struct Node n1 = {10, &n2};
  struct Node* head = &n1;

  struct Node* temp = head;
  while (temp != NULL) {
    printf("%d -> ", temp->data);
    temp = temp->next;
  }
  printf("NULL\n");
  return 0;
}

medium

A. 10 -> 20 -> 30 -> NULL

B. 30 -> 20 -> 10 -> NULL

C. 10 20 30 NULL

D. Compilation error

Solution

1. Step 1: Trace linked list nodes

   The list nodes are n1(10) -> n2(20) -> n3(30) -> NULL.

2. Step 2: Follow traversal and print

   The loop prints each node's data followed by ' -> ', then prints 'NULL' at the end.

3. Final Answer:

   10 -> 20 -> 30 -> NULL -> Option A

4. Quick Check:

   Print nodes in order = 10 -> 20 -> 30 -> NULL [OK]

Hint: Follow next pointers to print nodes in order [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing '->' arrows in output
• Forgetting to print NULL at end

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

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

void printList(struct Node* head) {
  while (head->next != NULL) {
    printf("%d -> ", head->data);
    head = head->next;
  }
  printf("NULL\n");
}

medium

A. head should be a double pointer

B. Missing semicolon after printf

C. printf format specifier is incorrect

D. The loop condition should be head != NULL

Solution

1. Step 1: Analyze loop condition

   The loop stops when head->next is NULL, so it skips printing the last node.

2. Step 2: Correct loop condition

   Changing condition to head != NULL ensures all nodes, including last, are printed.

3. Final Answer:

   The loop condition should be head != NULL -> Option D

4. Quick Check:

   Loop until head is NULL to print all nodes [OK]

Hint: Loop until pointer is NULL, not next pointer [OK]

Common Mistakes:

• Stopping loop too early
• Using wrong pointer type
• Ignoring last node in print

5. Given a linked list where some nodes may have data value 0, which of the following code snippets correctly prints all nodes including those with 0?

struct Node* temp = head;
while (temp) {
  if (temp->data) {
    printf("%d -> ", temp->data);
  }
  temp = temp->next;
}
printf("NULL\n");

hard

A. Change 'if (temp->data)' to always print without condition

B. Add 'temp = temp->next;' inside the if block only

C. Use 'while (temp != NULL)' and keep the if condition

D. Replace 'printf' with 'puts' to print zeros

Solution

1. Step 1: Understand the if condition effect

   The condition 'if (temp->data)' skips printing nodes with data 0 because 0 is false.

2. Step 2: Fix to print all nodes

   Removing the if condition prints all nodes including those with 0.

3. Final Answer:

   Change 'if (temp->data)' to always print without condition -> Option A

4. Quick Check:

   Print all nodes regardless of data value [OK]

Hint: Do not skip nodes with zero data when printing [OK]

Common Mistakes:

• Skipping zero values due to if condition
• Moving pointer only inside if block causing infinite loop
• Using wrong print function for integers