Insert at Beginning of Doubly Linked List in DSA Python

Mental Model

Add a new item at the start by linking it before the current first item and updating pointers.

Analogy: Imagine adding a new book at the front of a shelf where each book knows the one before and after it.

null ← 1 ↔ 2 ↔ 3 -> null

Dry Run Walkthrough

Input: list: 1 ↔ 2 ↔ 3, insert value 0 at beginning

Goal: Add 0 at the start so the list becomes 0 ↔ 1 ↔ 2 ↔ 3

Step 1: Create new node with value 0

null ← 1 ↔ 2 ↔ 3 -> null, new_node(0)

Why: We need a new node to insert at the beginning

Step 2: Set new_node.next to current head (1)

new_node(0) -> 1 ↔ 2 ↔ 3 -> null

Why: Link new node forward to old first node

Step 3: Set current head.prev to new_node

new_node(0) ↔ 1 ↔ 2 ↔ 3 -> null

Why: Link old first node back to new node

Step 4: Update head pointer to new_node

null ← 0 ↔ 1 ↔ 2 ↔ 3 -> null ↑head

Why: New node is now the first node in the list

Result:

null ← 0 ↔ 1 ↔ 2 ↔ 3 -> null

Annotated Code

DSA Python

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

class DoublyLinkedList:
    def __init__(self):
        self.head = None

    def insert_at_beginning(self, data):
        new_node = Node(data)  # create new node
        if self.head is None:
            self.head = new_node  # empty list, new node is head
            return
        new_node.next = self.head  # link new node forward to old head
        self.head.prev = new_node  # link old head back to new node
        self.head = new_node  # update head to new node

    def print_list(self):
        curr = self.head
        result = []
        while curr:
            result.append(str(curr.data))
            curr = curr.next
        print(" ↔ ".join(result) + " -> null")

# Driver code
dll = DoublyLinkedList()
dll.insert_at_beginning(3)
dll.insert_at_beginning(2)
dll.insert_at_beginning(1)
# List is now 1 ↔ 2 ↔ 3

# Insert 0 at beginning
dll.insert_at_beginning(0)
dll.print_list()

new_node = Node(data) # create new node

create new node to insert

if self.head is None: self.head = new_node # empty list, new node is head return

handle empty list by setting head to new node

new_node.next = self.head # link new node forward to old head

link new node's next to current head

self.head.prev = new_node # link old head back to new node

link old head's prev to new node

self.head = new_node # update head to new node

update head pointer to new node

Output

0 ↔ 1 ↔ 2 ↔ 3 -> null

Complexity Analysis

Time: O(1) because insertion at the beginning only changes a few pointers without traversal

Space: O(1) because no extra space is used except the new node

vs Alternative: Compared to inserting at the end which may require traversal O(n), this is faster and simpler

Edge Cases

empty list

new node becomes the head with no next or prev

DSA Python

if self.head is None:
            self.head = new_node  # empty list, new node is head
            return

single element list

new node links to old head, old head links back to new node

DSA Python

new_node.next = self.head  # link new node forward to old head

When to Use This Pattern

When you need to add an item quickly at the start of a list and keep two-way links, use insert at beginning of doubly linked list.

Common Mistakes

Mistake: Forgetting to update the old head's prev pointer to the new node
Fix: Always set old head.prev = new_node after linking new_node.next to old head

Mistake: Not handling empty list case causing errors
Fix: Check if head is None and set head to new node directly

Summary

Adds a new node at the start of a doubly linked list by updating next and prev pointers.

Use when you want to quickly add elements to the front of a list with backward and forward links.

The key is to link the new node before the current head and update the head pointer.

Practice

1. What is the main purpose of inserting a node at the beginning of a doubly linked list?

easy

A. To add a new node as the first element of the list

B. To remove the last node from the list

C. To reverse the entire list

D. To find the middle node of the list

Solution

1. Step 1: Understand the operation

   Inserting at the beginning means adding a new node before the current first node.

2. Step 2: Identify the purpose

   This operation makes the new node the first element in the doubly linked list.

3. Final Answer:

   To add a new node as the first element of the list -> Option A

4. Quick Check:

   Insert at beginning = Add new first node [OK]

Hint: Insert at beginning means new node becomes the first [OK]

Common Mistakes:

• Confusing insertion at beginning with deletion
• Thinking it reverses the list
• Assuming it finds a node instead of adding

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

easy

A. new_node = Node(data); new_node.prev = head; head.next = new_node; head = new_node

B. new_node = Node(data); head.next = new_node; new_node.prev = head; head = new_node

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

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

Solution

1. Step 1: Create new node and link next pointer

   new_node.next should point to current head to place it before head.

2. Step 2: Update previous head's prev pointer and head reference

   head.prev must point back to new_node, then update head to new_node.

3. Final Answer:

   new_node = Node(data); new_node.next = head; head.prev = new_node; head = new_node -> Option C

4. Quick Check:

   New node next = head, head prev = new node [OK]

Hint: New node next points to head; head prev points back [OK]

Common Mistakes:

• Swapping prev and next pointers
• Not updating head to new node
• Forgetting to update old head's prev pointer

3. What will be the output of the following code after inserting 10 at the beginning of the doubly linked list?

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

head = Node(20)
new_node = Node(10)
new_node.next = head
head.prev = new_node
head = new_node

# Print list forward
current = head
result = []
while current:
    result.append(current.data)
    current = current.next
print(result)

medium

A. [20, 10]

B. [10, 20]

C. [10]

D. [20]

Solution

1. Step 1: Insert 10 before 20

   New node with data 10 points next to old head (20), old head prev points back to new node.

2. Step 2: Traverse from new head and collect data

   Starting at head (10), next is 20, then None. So list is [10, 20].

3. Final Answer:

   [10, 20] -> Option B

4. Quick Check:

   List after insertion = [10, 20] [OK]

Hint: Traverse from head to next to see order [OK]

Common Mistakes:

• Confusing order and printing reversed list
• Not updating head pointer
• Printing only one node

4. Identify the error in the following code snippet that inserts a node at the beginning of a doubly linked list:

new_node = Node(5)
new_node.next = head
head = new_node
head.prev = new_node

medium

A. head.prev should be set before updating head

B. new_node.next should be None

C. head.prev should point to None, not new_node

D. No error, code is correct

Solution

1. Step 1: Analyze pointer assignments

   new_node.next points to old head (correct), but head = new_node happens before head.prev = new_node.

2. Step 2: Identify the error

   By the time head.prev = new_node executes, head is already the new_node, creating a self-loop. The old head's prev is never set. head.prev must be set before updating head to link the old head back correctly.

3. Final Answer:

   head.prev should be set before updating head -> Option A

4. Quick Check:

   Update old head.prev before head = new_node [OK]

Hint: Set head.prev before head = new_node [OK]

Common Mistakes:

• Setting prev pointer after head update incorrectly
• Making first node's prev point to itself
• Ignoring prev pointer for head node

5. Given a doubly linked list with nodes 30 <-> 40 <-> 50, write the sequence of node values after inserting 20 at the beginning and then 10 at the beginning. What is the final list?

hard

A. 10 <-> 30 <-> 20 <-> 40 <-> 50

B. 20 <-> 10 <-> 30 <-> 40 <-> 50

C. 30 <-> 40 <-> 50 <-> 20 <-> 10

D. 10 <-> 20 <-> 30 <-> 40 <-> 50

Solution

1. Step 1: Insert 20 at beginning

   New node 20 points next to 30, 30 prev points to 20. List becomes 20 <-> 30 <-> 40 <-> 50.

2. Step 2: Insert 10 at beginning

   New node 10 points next to 20, 20 prev points to 10. List becomes 10 <-> 20 <-> 30 <-> 40 <-> 50.

3. Final Answer:

   10 <-> 20 <-> 30 <-> 40 <-> 50 -> Option D

4. Quick Check:

   Insertions add nodes at front in order [OK]

Hint: Insertions at beginning stack nodes in reverse order [OK]

Common Mistakes:

• Inserting nodes in wrong order
• Appending instead of prepending
• Not updating prev pointers correctly