A doubly linked list is a chain of nodes where each node has three parts: data, a pointer to the next node, and a pointer to the previous node. This allows traversal in both directions.
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beginner
What steps are involved in inserting a node at the end of a doubly linked list?
1. Create a new node with the given data. 2. If the list is empty, set the new node as the head. 3. Otherwise, traverse to the last node. 4. Set the last node's next pointer to the new node. 5. Set the new node's previous pointer to the last node. 6. Set the new node's next pointer to null.
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beginner
Why do we need to update the previous pointer of the new node when inserting at the end?
Because the new node must know which node comes before it to maintain the doubly linked list structure, allowing backward traversal.
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intermediate
What happens if you forget to set the new node's next pointer to null when inserting at the end?
The list may have incorrect links, possibly causing traversal errors or infinite loops because the new node might point to an unintended node.
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beginner
Show the TypeScript code snippet to create a new node for a doubly linked list.
Hint: Node needs data, next, and prev properties [OK]
Common Mistakes:
Omitting prev pointer
Using wrong property names like value or nextNode
Setting next or prev as undefined instead of null
3. Given the following TypeScript code for inserting at the end of a doubly linked list, what will be the printed list after insertion?
class Node {
data: number;
next: Node | null = null;
prev: Node | null = null;
constructor(data: number) { this.data = data; }
}
class DoublyLinkedList {
head: Node | null = null;
insertAtEnd(data: number) {
const newNode = new Node(data);
if (!this.head) {
this.head = newNode;
return;
}
let temp = this.head;
while (temp.next) {
temp = temp.next;
}
temp.next = newNode;
newNode.prev = temp;
}
printList() {
let temp = this.head;
let result = '';
while (temp) {
result += temp.data + ' -> ';
temp = temp.next;
}
result += 'null';
console.log(result);
}
}
const dll = new DoublyLinkedList();
dll.insertAtEnd(10);
dll.insertAtEnd(20);
dll.insertAtEnd(30);
dll.printList();
medium
A. 10 -> 20 -> 30 -> null
B. 30 -> 20 -> 10 -> null
C. 10 -> 20 -> null
D. null
Solution
Step 1: Trace insertAtEnd calls
First insert adds 10 as head. Second adds 20 after 10. Third adds 30 after 20.
Step 2: Trace printList output
Print starts at head (10), then 20, then 30, then null.
Final Answer:
10 -> 20 -> 30 -> null -> Option A
Quick Check:
Inserted nodes in order = 10 -> 20 -> 30 [OK]
Hint: Insertion at end appends nodes in order [OK]
Common Mistakes:
Confusing order and printing reversed list
Missing last inserted node
Printing null instead of list
4. Identify the error in the following TypeScript method for inserting at the end of a doubly linked list:
insertAtEnd(data: number) {
const newNode = new Node(data);
if (!this.head) {
this.head = newNode;
}
let temp = this.head;
while (temp.next) {
temp = temp.next;
}
temp.next = newNode;
newNode.prev = temp;
}
medium
A. newNode should be created outside the method
B. Incorrect assignment of newNode.prev
C. Loop condition should be temp !== null
D. Missing return after setting head when list is empty
Solution
Step 1: Analyze empty list case
If head is null, method sets head to newNode but continues to loop and assign next, causing error.
Step 2: Fix by adding return after setting head
Adding return after setting head prevents further code execution and avoids null pointer errors.
Final Answer:
Missing return after setting head when list is empty -> Option D
Quick Check:
Return after head set avoids errors [OK]
Hint: Return immediately after setting head if list empty [OK]
Common Mistakes:
Not returning after setting head
Misunderstanding loop condition
Incorrect pointer assignments
5. You want to insert a new node with data 40 at the end of a doubly linked list that currently has nodes 10 <-> 20 <-> 30. After insertion, what will be the correct state of the list including the prev and next pointers of the new node?
hard
A. 10 <-> 20 <-> 30, with 40.prev and 40.next both null (not linked)
B. 40 <-> 10 <-> 20 <-> 30, where 40.next points to 10 and 40.prev is null
C. 10 <-> 20 <-> 30 <-> 40, where 40.prev points to 30 and 40.next is null
D. 10 <-> 20 <-> 30 <-> 40, where 40.prev and 40.next both point to null
Solution
Step 1: Understand insertion at end
The new node 40 should be linked after 30, so 30.next = 40 and 40.prev = 30.
Step 2: Confirm pointers of new node
Since 40 is last, 40.next must be null to mark list end.
Final Answer:
10 <-> 20 <-> 30 <-> 40, where 40.prev points to 30 and 40.next is null -> Option C
Quick Check:
New tail prev = old tail, next = null [OK]
Hint: New tail node prev = old tail, next = null [OK]
Common Mistakes:
Setting new node's next pointer incorrectly
Not linking new node's prev pointer
Inserting new node at start instead of end
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