Traversal and Printing a Linked List in DSA C

Mental Model

We start at the first item and move step-by-step to the next until we reach the end.

Analogy: Like reading a chain of paper clips one by one from the first to the last without skipping any.

head -> [1] -> [2] -> [3] -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3, print all values

Goal: Print all values in the list from start to end in order

Step 1: Start at the first node with value 1

head -> [1 ↑] -> [2] -> [3] -> null

Why: We begin printing from the first node

Step 2: Print value 1 and move to next node

head -> [1] -> [2 ↑] -> [3] -> null

Why: After printing current, move to next to continue

Step 3: Print value 2 and move to next node

head -> [1] -> [2] -> [3 ↑] -> null

Why: Continue printing each node in order

Step 4: Print value 3 and move to next node

head -> [1] -> [2] -> [3] -> null ↑

Why: Print last node and reach end (null)

Result:

head -> [1] -> [2] -> [3] -> null
Printed output: 1 2 3 

Annotated Code

DSA C

#include <stdio.h>
#include <stdlib.h>

// Define node structure
typedef struct Node {
    int data;
    struct Node* next;
} Node;

// Function to create a new node
Node* createNode(int value) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = value;
    newNode->next = NULL;
    return newNode;
}

// Function to print the linked list
void printList(Node* head) {
    Node* current = head; // Start at head
    while (current != NULL) { // Traverse until end
        printf("%d ", current->data); // Print current node's data
        current = current->next; // Move to next node
    }
    printf("\n");
}

int main() {
    // Create linked list 1 -> 2 -> 3 -> null
    Node* head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);

    // Print the list
    printList(head);

    // Free memory
    Node* temp;
    while (head != NULL) {
        temp = head;
        head = head->next;
        free(temp);
    }

    return 0;
}

Node* current = head; // Start at head

Initialize pointer to start traversal from first node

while (current != NULL) { // Traverse until end

Continue until we reach the end of the list

printf("%d ", current->data); // Print current node's data

Output the value stored in current node

current = current->next; // Move to next node

Advance pointer to next node to continue traversal

Output

1 2 3 

Complexity Analysis

Time: O(n) because we visit each of the n nodes exactly once

Space: O(1) because we only use a single pointer variable regardless of list size

vs Alternative: Compared to recursive printing which uses O(n) space on call stack, this iterative approach is more space efficient

Edge Cases

Empty list (head is NULL)

No output is printed, function exits immediately

DSA C

while (current != NULL) {

Single node list

Prints the single node's value and stops

DSA C

while (current != NULL) {

When to Use This Pattern

When you need to visit every item in a linked list in order, use traversal with a pointer moving from head to null.

Common Mistakes

Mistake: Not checking for NULL before accessing node data causing crashes
Fix: Always check current != NULL in the loop condition before accessing current->data

Mistake: Advancing pointer before printing causing skipped nodes
Fix: Print current node's data before moving pointer to next

Summary

It visits each node from start to end and prints its value.

Use it when you want to see or process all elements in a linked list.

Remember to move step-by-step from one node to the next until you reach the end.