Push Using Linked List Node in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'push' operation do in a linked list?

It adds a new node at the beginning (head) of the linked list, making it the new first element.

beginner

In C, what is the typical structure used to represent a node in a singly linked list?

A struct with at least two members: one for data (e.g., int data) and one pointer to the next node (e.g., struct Node* next).

intermediate

Why do we pass the head pointer by reference (pointer to pointer) when pushing a new node in C?

Because the head pointer itself may change to point to the new node, so we need to modify the original pointer outside the function.

beginner

What is the time complexity of the push operation in a singly linked list?

O(1) — It takes constant time because we only change a few pointers without traversing the list.

beginner

Show the basic steps to push a new node with value 'x' onto a linked list in C.

1. Create a new node using malloc. 2. Set new node's data to x. 3. Set new node's next to current head. 4. Update head to new node.

What pointer type should be passed to a push function to update the head of a linked list in C?

A Pointer to data (int*)

B Pointer to node (Node*)

C Pointer to pointer to node (Node**)

D Pointer to void (void*)

To update the head pointer itself, the function needs a pointer to the head pointer, which is Node**.

What happens to the old head node when a new node is pushed onto the linked list?

A It becomes the second node in the list

B It is deleted

C It becomes the last node

D It remains the head

The old head node is linked after the new node, becoming the second node.

Which of these is NOT a step in pushing a node onto a linked list?

A Setting new node's next to NULL always

B Allocating memory for the new node

C Updating head to point to the new node

D Assigning data to the new node

The new node's next should point to the old head, not always NULL.

What is the output linked list after pushing nodes with values 3, then 5, then 7 onto an empty list?

A 3 -> 7 -> 5 -> null

B 3 -> 5 -> 7 -> null

C 5 -> 3 -> 7 -> null

D 7 -> 5 -> 3 -> null

Each new node is added at the front, so the last pushed (7) is the head.

What is the time complexity of pushing a node onto a singly linked list?

A O(n)

B O(1)

C O(log n)

D O(n^2)

Push operation takes constant time because it only changes pointers at the head.

Explain how the push operation works in a singly linked list using a linked list node in C.

Think about how the new node becomes the first element.

Describe why the head pointer must be passed as a pointer to pointer when implementing push in C.

Consider how C passes arguments by value.

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