Enqueue Using Linked List in DSA Python

Mental Model

Add new items to the end of a line by linking them one after another.

Analogy: Imagine people standing in a queue where each person holds the hand of the next person. When a new person arrives, they join at the end by holding the last person's hand.

front -> null
rear -> null

(empty queue with no nodes)

Dry Run Walkthrough

Input: Start with empty queue, enqueue values 10, then 20, then 30

Goal: Add each new value to the end of the linked list representing the queue

Step 1: enqueue 10 into empty queue

front -> [10] -> null
rear -> [10] -> null

Why: First node becomes both front and rear because queue was empty

Step 2: enqueue 20 at the end

front -> [10] -> [20] -> null
rear -> [20] -> null

Why: Link new node after rear and update rear pointer to new node

Step 3: enqueue 30 at the end

front -> [10] -> [20] -> [30] -> null
rear -> [30] -> null

Why: Link new node after rear and update rear pointer to new node

Result:

front -> 10 -> 20 -> 30 -> null
rear -> 30 -> null

Annotated Code

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class Queue:
    def __init__(self):
        self.front = None
        self.rear = None

    def enqueue(self, value):
        new_node = Node(value)
        if self.rear is None:
            self.front = new_node
            self.rear = new_node
            return
        self.rear.next = new_node
        self.rear = new_node

    def __str__(self):
        result = []
        current = self.front
        while current:
            result.append(str(current.data))
            current = current.next
        return ' -> '.join(result) + ' -> null'

# Driver code
queue = Queue()
queue.enqueue(10)
queue.enqueue(20)
queue.enqueue(30)
print(queue)

if self.rear is None:

check if queue is empty to set front and rear to new node

self.front = new_node self.rear = new_node

initialize front and rear pointers to the first node

self.rear.next = new_node

link new node after current rear node

self.rear = new_node

update rear pointer to new node

Output

10 -> 20 -> 30 -> null

Complexity Analysis

Time: O(1) because enqueue adds node directly at rear without traversal

Space: O(n) because each enqueue adds one new node to the linked list

vs Alternative: Compared to array-based queue, linked list enqueue avoids shifting elements and is always O(1)

Edge Cases

enqueue on empty queue

front and rear both point to the new node

DSA Python

if self.rear is None:

enqueue multiple elements

rear pointer moves forward each time, front stays at first node

DSA Python

self.rear = new_node

When to Use This Pattern

When you need to add items to the end of a queue efficiently, use enqueue with a linked list to keep O(1) insertion time.

Common Mistakes

Mistake: Not updating rear pointer after adding new node
Fix: Always set rear to new_node after linking it

Mistake: Not handling empty queue case separately
Fix: Check if rear is None and set front and rear to new_node

Summary

Adds new elements to the end of a queue using a linked list.

Use when you want efficient O(1) insertion at the queue's rear.

Remember to update rear pointer and handle empty queue case carefully.

Practice

1. What does the enqueue operation do in a linked list implementation of a queue?

easy

A. Adds a new element at the end (tail) of the linked list

B. Removes the first element from the linked list

C. Adds a new element at the beginning (head) of the linked list

D. Deletes the last element from the linked list

Solution

1. Step 1: Understand enqueue operation in queue

   Enqueue means adding an element to the queue, which is done at the end (tail) to maintain FIFO order.

2. Step 2: Linked list enqueue behavior

   In a linked list, adding at the tail means creating a new node and linking it as the last node.

3. Final Answer:

   Adds a new element at the end (tail) of the linked list -> Option A

4. Quick Check:

   Enqueue = Add at tail [OK]

Hint: Enqueue always adds at the tail in linked list queue [OK]

Common Mistakes:

• Confusing enqueue with dequeue operation
• Adding new element at head instead of tail
• Thinking enqueue removes elements

2. Which of the following is the correct syntax to create a new node and enqueue it at the tail in Python linked list?

easy

A. new_node = Node(data); new_node.next = tail; tail = new_node

B. new_node = Node(data); tail.next = new_node; tail = new_node

C. new_node = Node(data); new_node.next = head; head = new_node

D. new_node = Node(data); head.next = new_node; head = new_node

Solution

1. Step 1: Create new node with data

   We create a new node object with the given data.

2. Step 2: Link new node at tail and update tail pointer

   Set current tail's next to new node, then update tail to new node.

3. Final Answer:

   new_node = Node(data); tail.next = new_node; tail = new_node -> Option B

4. Quick Check:

   New node linked at tail [OK]

Hint: Tail.next points to new node, then tail moves to new node [OK]

Common Mistakes:

• Linking new node to head instead of tail
• Not updating tail pointer after enqueue
• Setting new_node.next to tail (wrong direction)

3. What will be the state of the queue after executing this code?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class Queue:
    def __init__(self):
        self.head = None
        self.tail = None

    def enqueue(self, data):
        new_node = Node(data)
        if not self.tail:
            self.head = self.tail = new_node
        else:
            self.tail.next = new_node
            self.tail = new_node

q = Queue()
q.enqueue(10)
q.enqueue(20)
q.enqueue(30)
current = q.head
result = []
while current:
    result.append(current.data)
    current = current.next
print(result)

medium

A. []

B. [30, 20, 10]

C. [10, 30, 20]

D. [10, 20, 30]

Solution

1. Step 1: Enqueue 10, 20, 30 in order

   First enqueue sets head and tail to node with 10. Next enqueues add nodes at tail with 20 and 30.

2. Step 2: Traverse from head to tail collecting data

   Traversal collects 10, then 20, then 30 in order.

3. Final Answer:

   [10, 20, 30] -> Option D

4. Quick Check:

   Queue order = insertion order [OK]

Hint: Enqueue adds at tail, so order stays insertion order [OK]

Common Mistakes:

• Assuming enqueue adds at head reversing order
• Not initializing head and tail properly
• Confusing output list order

4. Find the error in this enqueue method for a linked list queue:

def enqueue(self, data):
    new_node = Node(data)
    if self.head is None:
        self.head = new_node
    else:
        self.tail.next = new_node
        self.tail = new_node

medium

A. Does not set new_node.next to None

B. Does not update head when queue is empty

C. Does not update tail when queue is empty

D. Does not link new_node to tail.next

Solution

1. Step 1: Check initialization when queue is empty

   When head is None, head is set to new_node but tail is not updated.

2. Step 2: Tail must also point to new_node when queue is empty

   Without updating tail, tail remains None causing errors on next enqueue.

3. Final Answer:

   Does not update tail when queue is empty -> Option C

4. Quick Check:

   Empty queue needs both head and tail set [OK]

Hint: Set both head and tail on empty queue enqueue [OK]

Common Mistakes:

• Updating only head or only tail on empty queue
• Forgetting to set tail.next on non-empty queue
• Assuming new_node.next needs manual None assignment

5. You want to implement a queue using a linked list that supports enqueue and dequeue in O(1) time. Which of the following changes is necessary to the enqueue method to handle the edge case when the queue is empty?

hard

A. Set both head and tail to the new node when queue is empty

B. Set only tail to the new node when queue is empty

C. Set new node's next to head when queue is empty

D. Set tail.next to head when queue is empty

Solution

1. Step 1: Understand queue empty state

   When queue is empty, both head and tail pointers are None.

2. Step 2: Enqueue first element must update both pointers

   To maintain O(1) enqueue and dequeue, head and tail must both point to the new node.

3. Step 3: This ensures correct linking and prevents errors

   Without setting both, dequeue or next enqueue operations may fail or cause errors.

4. Final Answer:

   Set both head and tail to the new node when queue is empty -> Option A

5. Quick Check:

   Empty queue enqueue sets head and tail [OK]

Hint: Empty queue needs head and tail set to new node [OK]

Common Mistakes:

• Setting only tail or only head on empty enqueue
• Linking new node incorrectly to head or tail
• Not handling empty queue edge case