Traversal and Printing a Linked List in DSA C - Execution Trace

Concept Flow - Traversal and Printing a Linked List

Start at head node

↓

Is current node NULL?

Yes→Stop traversal

No↓

Print current node data

↓

Move to next node

↩Back to check current node

Start from the first node, print its data, move to the next node, and repeat until no nodes remain.

Execution Sample

DSA C

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

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

This code walks through each node in the linked list from head to end, printing the data values.

Execution Table

Step	Operation	Current Node Data	Pointer Changes	Visual State
1	Start at head	10	current = head (node with 10)	10 -> 20 -> 30 -> NULL
2	Print current node data	10	No pointer change	10 -> 20 -> 30 -> NULL
3	Move to next node	20	current = current->next (node with 20)	10 -> 20 -> 30 -> NULL
4	Print current node data	20	No pointer change	10 -> 20 -> 30 -> NULL
5	Move to next node	30	current = current->next (node with 30)	10 -> 20 -> 30 -> NULL
6	Print current node data	30	No pointer change	10 -> 20 -> 30 -> NULL
7	Move to next node	NULL	current = current->next (NULL)	10 -> 20 -> 30 -> NULL
8	Check current node	NULL	current == NULL, stop traversal	10 -> 20 -> 30 -> NULL

💡 Traversal stops when current pointer becomes NULL, meaning end of list reached.

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	Final
current	head (node with 10)	node with 20	node with 30	NULL	NULL

Key Moments - 3 Insights

Why do we check if current is NULL before printing?

Because if current is NULL, it means we reached the end of the list and there is no data to print. This is shown in execution_table step 8 where traversal stops.

Why do we move current to current->next after printing?

To advance to the next node in the list so we can print its data next. This is shown in steps 3, 5, and 7 where current pointer moves forward.

What does the visual state '10 -> 20 -> 30 -> NULL' represent?

It shows the linked list nodes and their order. The arrows mean each node points to the next, ending with NULL meaning no more nodes.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of current after step 5?

A NULL

B Node with data 20

C Node with data 30

D Node with data 10

💡 Hint

Check the 'Pointer Changes' and 'Current Node Data' columns at step 5 in execution_table.

At which step does the traversal stop because current becomes NULL?

A Step 6

B Step 8

C Step 7

D Step 5

💡 Hint

Look at the 'Operation' and 'Pointer Changes' columns in execution_table for when current == NULL.

If the list had only one node with data 5, how many print operations would occur?

A 1

B 0

C 2

D 3

💡 Hint

Refer to variable_tracker and execution_table logic for how many times printing happens per node.

Concept Snapshot

Traversal and Printing a Linked List:
- Start at head node
- While current node is not NULL:
  - Print current node data
  - Move to next node
- Stop when current is NULL
- Output format: data -> data -> ... -> NULL

Full Transcript

Traversal and printing a linked list means starting at the first node and moving through each node one by one. At each node, we print its data. We keep moving to the next node until we reach the end, which is marked by a NULL pointer. The code uses a pointer called current to track which node we are at. We check if current is NULL before printing to avoid errors. After printing, we move current to the next node. This repeats until current becomes NULL, which stops the traversal. The linked list is shown visually as nodes connected by arrows ending with NULL.

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