Push Using Linked List Node in DSA C

Mental Model

Add a new item at the start of a linked list by making it point to the current first item.

Analogy: Imagine adding a new book on top of a stack of books; the new book becomes the first one you see.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, push value 0

Goal: Add 0 at the start so the list becomes 0 -> 1 -> 2 -> 3 -> null

Step 1: Create new node with value 0

new_node[0] -> null, head -> 1 -> 2 -> 3 -> null

Why: We need a new node to add to the list

Step 2: Set new_node's next to current head (node with value 1)

new_node[0] -> 1 -> 2 -> 3 -> null, head -> 1 -> 2 -> 3 -> null

Why: Link new node to the existing list so it points to the first node

Step 3: Update head to new_node

head -> 0 -> 1 -> 2 -> 3 -> null

Why: New node becomes the first node in the list

Result:

head -> 0 -> 1 -> 2 -> 3 -> null

Annotated Code

DSA C

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

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

// Push function to add node at start
void push(struct Node** head_ref, int new_data) {
    // Create new node
    struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
    new_node->data = new_data;

    // Link new node to current head
    new_node->next = *head_ref;

    // Update head to new node
    *head_ref = new_node;
}

// Function to print list
void printList(struct Node* node) {
    while (node != NULL) {
        printf("%d -> ", node->data);
        node = node->next;
    }
    printf("null\n");
}

int main() {
    // Start with list 1->2->3->null
    struct Node* head = NULL;

    // Push 3
    push(&head, 3);
    // Push 2
    push(&head, 2);
    // Push 1
    push(&head, 1);

    // Print original list
    printList(head);

    // Push 0 at start
    push(&head, 0);

    // Print updated list
    printList(head);

    return 0;
}

struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));

Create new node to hold new data

new_node->next = *head_ref;

Link new node to current head node

*head_ref = new_node;

Update head pointer to new node

Output

1 -> 2 -> 3 -> null

0 -> 1 -> 2 -> 3 -> null

Complexity Analysis

Time: O(1) because we only change pointers without traversing the list

Space: O(1) because we allocate one new node regardless of list size

vs Alternative: Compared to adding at the end which is O(n), pushing at start is faster and simpler

Edge Cases

Empty list (head is NULL)

New node becomes the only node and head points to it

DSA C

new_node->next = *head_ref;

When to Use This Pattern

When you need to add an element quickly at the start of a linked list, use push to update head pointer.

Common Mistakes

Mistake: Not updating the head pointer after creating the new node
Fix: Assign the new node to head pointer to make it the first node

Mistake: Forgetting to link new node's next to the old head
Fix: Set new_node->next to current head before updating head

Summary

Adds a new node at the beginning of a linked list.

Use when you want to insert elements quickly at the start.

The new node points to the old first node, then head updates to new node.

Practice

1. What does the push operation do in a linked list when using nodes?

easy

A. Searches for a node with a specific value

B. Adds a new node at the beginning of the list

C. Removes the last node from the list

D. Deletes the entire linked list

Solution

1. Step 1: Understand the push operation

   The push operation adds a new element to the linked list.

2. Step 2: Identify where the new node is added

   In a linked list, push adds the new node at the start (head) of the list.

3. Final Answer:

   Adds a new node at the beginning of the list -> Option B

4. Quick Check:

   Push = Add at head [OK]

Hint: Push always adds at the start of the linked list [OK]

Common Mistakes:

• Thinking push adds at the end
• Confusing push with pop
• Assuming push deletes nodes

2. Which of the following is the correct syntax to create a new node and push it onto the linked list in C?

easy

A. newNode = malloc(sizeof(Node)); newNode->prev = head; head = newNode;

B. newNode = malloc(Node); newNode->next = head; head = newNode;

C. newNode = malloc(sizeof(Node)); newNode->next = head; head = newNode;

D. newNode = malloc(sizeof(Node)); head = newNode->next; newNode = head;

Solution

1. Step 1: Allocate memory correctly for new node

   Use malloc(sizeof(Node)) to allocate memory for the new node.

2. Step 2: Link new node to current head and update head

   Set newNode->next to current head, then update head to newNode.

3. Final Answer:

   newNode = malloc(sizeof(Node)); newNode->next = head; head = newNode; -> Option C

4. Quick Check:

   Correct malloc and linking = newNode = malloc(sizeof(Node)); newNode->next = head; head = newNode; [OK]

Hint: Use sizeof(Node) with malloc and link newNode->next to head [OK]

Common Mistakes:

• Forgetting sizeof in malloc
• Assigning head before setting newNode->next
• Using prev pointer in singly linked list

3. What will be the linked list after executing the following code?

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

Node* head = NULL;

void push(int val) {
  Node* newNode = malloc(sizeof(Node));
  newNode->data = val;
  newNode->next = head;
  head = newNode;
}

push(10);
push(20);
push(30);

medium

A. 30 -> 20 -> 10 -> null

B. 10 -> 20 -> 30 -> null

C. 20 -> 30 -> 10 -> null

D. 10 -> 30 -> 20 -> null

Solution

1. Step 1: Trace each push call

   push(10) adds 10 at head: 10 -> null; push(20) adds 20 at head: 20 -> 10 -> null; push(30) adds 30 at head: 30 -> 20 -> 10 -> null.

2. Step 2: Confirm final linked list order

   The last pushed value is at the head, so list is 30 -> 20 -> 10 -> null.

3. Final Answer:

   30 -> 20 -> 10 -> null -> Option A

4. Quick Check:

   Last pushed is head = 30 -> 20 -> 10 [OK]

Hint: Last pushed node is always at the head [OK]

Common Mistakes:

• Assuming push adds at tail
• Reading list in wrong order
• Confusing push with append

4. Identify the error in this push function and select the correct fix:

void push(int val) {
  Node* newNode;
  newNode->data = val;
  newNode->next = head;
  head = newNode;
}

medium

A. head should be assigned before newNode->next

B. newNode->data should be assigned after head update

C. newNode->next should be NULL instead of head

D. newNode is not allocated memory; fix by malloc(sizeof(Node))

Solution

1. Step 1: Check newNode initialization

   newNode is declared but not allocated memory with malloc, so it points to garbage.

2. Step 2: Fix by allocating memory

   Use newNode = malloc(sizeof(Node)) before accessing newNode->data or newNode->next.

3. Final Answer:

   newNode is not allocated memory; fix by malloc(sizeof(Node)) -> Option D

4. Quick Check:

   Always malloc before using newNode [OK]

Hint: Always allocate memory before using new node [OK]

Common Mistakes:

• Using uninitialized pointer
• Assigning data before allocation
• Changing head before linking newNode

5. You want to push multiple values 5, 15, 25 onto an empty linked list using the push function below. After pushing, you want to print the list from head to tail. Which code snippet correctly performs this and prints the list?

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

Node* head = NULL;

void push(int val) {
  Node* newNode = malloc(sizeof(Node));
  newNode->data = val;
  newNode->next = head;
  head = newNode;
}

// Your code here to push and print

hard

A. push(5); push(15); push(25); Node* temp = head; while(temp) { printf("%d -> ", temp->data); temp = temp->next; } printf("null\n");

B. push(5); push(15); push(25); Node* temp = head; while(temp->next) { printf("%d -> ", temp->data); temp = temp->next; } printf("null\n");

C. push(25); push(15); push(5); Node* temp = head; while(temp) { printf("%d -> ", temp->data); temp = temp->next; } printf("null\n");

D. push(5); push(15); push(25); Node* temp = head; while(temp) { printf("%d ", temp->data); temp = temp->next; }

Solution

1. Step 1: Push values in correct order

   Push 5, then 15, then 25. Each push adds at head, so list is 25 -> 15 -> 5 -> null.

2. Step 2: Print all nodes correctly

   Use a loop while(temp) to print each node's data followed by '->', then print 'null' after loop.

3. Final Answer:

   push(5); push(15); push(25); Node* temp = head; while(temp) { printf("%d -> ", temp->data); temp = temp->next; } printf("null\n"); -> Option A

4. Quick Check:

   Push order + full print loop = push(5); push(15); push(25); Node* temp = head; while(temp) { printf("%d -> ", temp->data); temp = temp->next; } printf("null\n"); [OK]

Hint: Push in order, print with while(temp) loop ending with null [OK]

Common Mistakes:

• Pushing in reverse order
• Using while(temp->next) and missing last node
• Not printing 'null' at end