Queue Implementation Using Linked List in DSA C++ - Execution Trace

Concept Flow - Queue Implementation Using Linked List

Create new node with data

↓

Is queue empty?

Yes→Set head and tail to new node

| No↓

Set tail.next to new node

↓

Update tail to new node

↓

Done

↓

Dequeue operation

↓

Is queue empty?

Yes→Return empty message

| No↓

Save head data

↓

Move head to head.next

↓

If head is null, set tail to null

↓

Return saved data

↓

Done

This flow shows how nodes are added at the tail and removed from the head in a linked list queue.

Execution Sample

DSA C++

enqueue(10)
enqueue(20)
dequeue()
enqueue(30)
dequeue()

This code adds 10 and 20 to the queue, removes one element, adds 30, then removes another element.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	enqueue(10)	10	head -> 10, tail -> 10	10 -> null
2	enqueue(20)	10 -> 20	tail.next -> 20, tail -> 20	10 -> 20 -> null
3	dequeue()	20	head -> 20	20 -> null
4	enqueue(30)	20 -> 30	tail.next -> 30, tail -> 30	20 -> 30 -> null
5	dequeue()	30	head -> 30	30 -> null
6	End	30	No change	30 -> null

💡 No more operations; queue has one element left.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
head	null	Node(10)	Node(10)	Node(20)	Node(20)	Node(30)	Node(30)
tail	null	Node(10)	Node(20)	Node(20)	Node(30)	Node(30)	Node(30)
size	0	1	2	1	2	1	1

Key Moments - 3 Insights

Why do we update tail.next when enqueueing but not head?

Because new nodes are added at the tail end, so tail.next points to the new node. Head stays the same until dequeue.

What happens when we dequeue the last element?

After removing the last node, head becomes null and we also set tail to null to mark the queue as empty.

Why does the visual state show '-> null' at the end?

It shows the end of the linked list. The last node's next pointer is null, meaning no more nodes after it.

Visual Quiz - 3 Questions

Look at the execution table, what is the visual state after step 3?

A 20 -> null

B 10 -> 20 -> null

C 30 -> null

D null

💡 Hint

Check the 'Visual State' column in row with Step 3.

At which step does the tail pointer move to Node(30)?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at 'Pointer Changes' column to see when tail updates to 30.

If we dequeue twice at the start, what would be the size after step 3?

A 1

B Cannot dequeue twice

C 0

D 2

💡 Hint

Refer to 'size' variable in variable_tracker and dequeue operation rules.

Concept Snapshot

Queue using linked list:
- Enqueue adds node at tail
- Dequeue removes node from head
- Update head and tail pointers carefully
- Empty queue: head and tail are null
- Visual: nodes linked with next pointers ending in null

Full Transcript

This visual trace shows how a queue is implemented using a linked list. Enqueue adds new nodes at the tail, updating tail.next and tail pointer. Dequeue removes nodes from the head, moving head pointer forward. When the last node is dequeued, both head and tail become null to mark the queue empty. The linked nodes are shown with arrows ending in null to indicate the list end. Variables head, tail, and size track the queue state step-by-step.

Practice

1. What is the main advantage of implementing a queue using a linked list instead of an array?

easy

A. Dynamic size without fixed capacity

B. Faster access to middle elements

C. Uses less memory always

D. Simpler to implement than arrays

Solution

1. Step 1: Understand queue size limitation in arrays

   Arrays have fixed size, so queues implemented with arrays have a maximum capacity.

2. Step 2: Linked list dynamic size

   Linked lists can grow or shrink dynamically, so queues using linked lists do not have fixed capacity limits.

3. Final Answer:

   Dynamic size without fixed capacity -> Option A

4. Quick Check:

   Linked list queues grow dynamically = D [OK]

Hint: Linked list queues grow dynamically, arrays don't [OK]

Common Mistakes:

• Thinking linked lists use less memory always
• Confusing access speed advantages
• Assuming linked lists are simpler to implement

2. Which of the following is the correct way to declare a node structure for a queue using a linked list in C++?

easy

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

B. struct Node { int data; Node next; };

C. struct Node { int data; Node* prev; };

D. struct Node { int data; int next; };

Solution

1. Step 1: Understand node structure for linked list

   A node must contain data and a pointer to the next node.

2. Step 2: Analyze options

   struct Node { int data; Node* next; }; correctly uses a pointer to next node. struct Node { int data; Node next; }; uses next as object, which is invalid. struct Node { int data; Node* prev; }; uses prev pointer, which is for doubly linked list. struct Node { int data; int next; }; uses int for next, which is incorrect.

3. Final Answer:

   struct Node { int data; Node* next; }; -> Option A

4. Quick Check:

   Node with data and next pointer = A [OK]

Hint: Node must have pointer to next node, not object or int [OK]

Common Mistakes:

• Using object instead of pointer for next
• Confusing singly and doubly linked list nodes
• Using int instead of pointer for next

3. What will be the output after performing these operations on an empty queue implemented using linked list?

enqueue(10); enqueue(20); dequeue(); enqueue(30); dequeue();

Print the queue elements from front to rear.

medium

A. 20 -> 30 -> null

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

C. 30 -> null

D. 10 -> 30 -> null

Solution

1. Step 1: Perform enqueue(10) and enqueue(20)

   Queue becomes: 10 -> 20 -> null

2. Step 2: Perform dequeue()

   Removes 10, queue becomes: 20 -> null

3. Step 3: Perform enqueue(30)

   Queue becomes: 20 -> 30 -> null

4. Step 4: Perform dequeue()

   Removes 20, queue becomes: 30 -> null

5. Final Answer:

   30 -> null -> Option C

6. Quick Check:

   After operations queue front is 30 [OK]

Hint: Track enqueue/dequeue stepwise to find final queue [OK]

Common Mistakes:

• Forgetting to remove front on dequeue
• Appending dequeue element instead of removing
• Mixing order of operations

4. Identify the error in this dequeue function for a queue implemented using linked list:

int dequeue() {
  if (front == nullptr) return -1;
  Node* temp = front;
  front = front->next;
  int val = temp->data;
  delete temp;
  return val;
}

medium

A. No error, function is correct

B. Does not update rear when queue becomes empty

C. Deletes front node before accessing data

D. Returns wrong value after deletion

Solution

1. Step 1: Analyze dequeue steps

   Function removes front node and returns its data correctly.

2. Step 2: Check rear pointer update

   If after dequeue front becomes nullptr (empty queue), rear should also be set to nullptr but this is missing.

3. Final Answer:

   Does not update rear when queue becomes empty -> Option B

4. Quick Check:

   Rear must be nullptr if queue empty [OK]

Hint: Update rear to nullptr when queue becomes empty [OK]

Common Mistakes:

• Ignoring rear pointer update on empty queue
• Deleting node before reading data
• Returning wrong value after deletion

5. You want to implement a function reverseQueue() that reverses the elements of a queue implemented using a linked list without using extra arrays or STL containers. Which approach is best?

hard

A. Use recursion to dequeue elements and enqueue them back in reverse order

B. Traverse linked list and swap data of front and rear nodes repeatedly

C. Create a new linked list and insert elements at front while dequeuing

D. Use two pointers to reverse links between nodes iteratively

Solution

1. Step 1: Understand reversing linked list nodes

   Reversing links between nodes iteratively changes the queue order without extra space.

2. Step 2: Analyze options

   Use recursion to dequeue elements and enqueue them back in reverse order uses recursion but may use call stack (extra space). Traverse linked list and swap data of front and rear nodes repeatedly swaps data but is inefficient and complex. Create a new linked list and insert elements at front while dequeuing creates new list (extra space). Use two pointers to reverse links between nodes iteratively reverses links in place efficiently.

3. Final Answer:

   Use two pointers to reverse links between nodes iteratively -> Option D

4. Quick Check:

   In-place link reversal is best for linked list queue reverse [OK]

Hint: Reverse links in place with two pointers for best solution [OK]

Common Mistakes:

• Using extra arrays or containers
• Swapping data instead of links
• Using recursion causing extra space