Why Push Using Linked List Node in DSA C?

The Big Idea

What if you could add items instantly without moving everything else?

The Scenario

Imagine you have a stack of books on your desk. You want to add a new book on top, but you have to move all the books one by one to make space. This is like manually shifting elements in an array to add a new item at the start.

The Problem

Manually shifting all elements to add a new item at the beginning is slow and tiring. It takes a lot of time as the stack grows, and you might make mistakes moving the books around. This is inefficient and error-prone.

The Solution

Using a linked list node to push means you just create a new node and link it to the current top. No shifting needed! It's like placing the new book directly on top without moving others. This makes adding fast and simple.

Before vs After

✗ Before

for (int i = size; i > 0; i--) {
  array[i] = array[i - 1];
}
array[0] = new_value;

✓ After

new_node->next = head;
head = new_node;

What It Enables

This lets you add items instantly at the start, making stacks and queues efficient and easy to manage.

Real Life Example

Think of a browser's back button history. Each new page you visit is pushed on top instantly, so you can go back step-by-step without delay.

Key Takeaways

Manual shifting is slow and error-prone.

Push with linked list node adds instantly at the start.

Efficient for stacks, queues, and history tracking.

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