Traversal and Printing a Linked List in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start traversal from the head node.

DSA Typescript

let current = [1];

Drag options to blanks, or click blank then click option'

Atail

Bnext

Cnull

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using tail instead of head to start traversal.

Assigning null directly which means no traversal.

Traversal of a linked list starts from the head node.

2fill in blank

medium

Complete the code to move to the next node during traversal.

DSA Typescript

current = current[1];

Drag options to blanks, or click blank then click option'

Atail

Bnext

Chead

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

Using 'head' or 'tail' which are not properties of the current node.

To move forward in a singly linked list, assign current to current.next.

3fill in blank

hard

Fix the error in the loop condition to traverse until the end of the list.

DSA Typescript

while (current [1] null) {
  console.log(current.value);
  current = current.next;
}

Drag options to blanks, or click blank then click option'

A!=

B==

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which stops immediately if current is null.

Using '<' or '>' which are not valid for object comparisons.

The loop should continue while current is not null, so use !=.

4fill in blank

hard

Fill both blanks to create a function that prints all values in the linked list.

DSA Typescript

function printList([1]: ListNode | null): void {
  let current = [2];
  while (current != null) {
    console.log(current.value);
    current = current.next;
  }
}

Drag options to blanks, or click blank then click option'

Ahead

Bnode

Ccurrent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'head' when the parameter is named differently.

Assigning 'current' to itself before initialization.

The function takes a parameter named node and starts traversal from it by assigning it to current.

5fill in blank

hard

Fill all three blanks to create a function that collects values from the linked list into an array.

DSA Typescript

function listValues([1]: ListNode | null): number[] {
  const values: number[] = [];
  let [2] = [3];
  while (current != null) {
    values.push(current.value);
    current = current.next;
  }
  return values;
}

Drag options to blanks, or click blank then click option'

Ahead

Bnode

Ccurrent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'head' as parameter but 'node' in code or vice versa.

Not initializing current to the parameter node.

The function parameter is node. We declare current and assign it to node to start traversal.

Practice

1. What does it mean to traverse a linked list?

easy

A. Visit each node in the list one by one from start to end

B. Add a new node at the end of the list

C. Delete the first node of the list

D. Reverse the order of nodes in the list

Solution

1. Step 1: Understand traversal meaning

   Traversal means going through each element in order to access or process it.

2. Step 2: Apply to linked list context

   In a linked list, traversal means visiting each node starting from the head until the end (null).

3. Final Answer:

   Visit each node in the list one by one from start to end -> Option A

4. Quick Check:

   Traversal = Visit nodes sequentially [OK]

Hint: Traversal means visiting nodes one after another [OK]

Common Mistakes:

• Confusing traversal with insertion or deletion
• Thinking traversal changes the list order
• Assuming traversal skips nodes

2. Which of the following is the correct way to start traversing a linked list in TypeScript?

easy

A. let current = head;

B. let current = tail;

C. let current = null;

D. let current = new Node();

Solution

1. Step 1: Identify starting point for traversal

   Traversal starts from the first node, called head, to visit all nodes.

2. Step 2: Check options for correct initialization

   Setting current to head allows moving through nodes; tail or null won't start traversal properly.

3. Final Answer:

   let current = head; -> Option A

4. Quick Check:

   Start traversal at head node [OK]

Hint: Always start traversal from the head node [OK]

Common Mistakes:

• Starting traversal from tail which is the last node
• Initializing current to null which skips traversal
• Creating a new node instead of using existing head

3. What will be the output of the following TypeScript code that prints a linked list?

class Node {
  value: number;
  next: Node | null;
  constructor(value: number) {
    this.value = value;
    this.next = null;
  }
}

const head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);

let current = head;
while (current !== null) {
  console.log(current.value);
  current = current.next;
}

medium

A. 3 2 1

B. 1 2 3

C. 1 3 2

D. 2 3 1

Solution

1. Step 1: Understand list structure

   The list nodes are linked as 1 -> 2 -> 3 -> null.

2. Step 2: Trace the while loop output

   The loop prints current.value starting at head (1), then moves to 2, then 3, then stops at null.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Print nodes in order: 1 2 3 [OK]

Hint: Print values while moving from head to next until null [OK]

Common Mistakes:

• Printing in reverse order by mistake
• Skipping nodes by incorrect pointer update
• Confusing node values order

4. Identify the error in this linked list traversal code and select the fix:

let current = head;
while (current.next !== null) {
  console.log(current.value);
  current = current.next;
}

medium

A. Remove console.log inside the loop

B. Add console.log after the loop

C. Initialize current to null

D. Change condition to current !== null

Solution

1. Step 1: Analyze loop condition

   The loop stops when current.next is null, so it skips printing the last node.

2. Step 2: Fix condition to include last node

   Changing condition to current !== null ensures all nodes, including last, are printed.

3. Final Answer:

   Change condition to current !== null -> Option D

4. Quick Check:

   Loop until current is null to print all nodes [OK]

Hint: Loop until current is null to include last node [OK]

Common Mistakes:

• Using current.next !== null misses last node
• Not updating current inside loop causes infinite loop
• Printing after loop instead of inside

5. Given a linked list with nodes containing values 5 -> 10 -> 15 -> null, write a TypeScript function to traverse and collect these values into an array. What will be the returned array after traversal?

hard

A. [15, 10, 5]

B. [5, 15, 10]

C. [5, 10, 15]

D. [10, 5, 15]

Solution

1. Step 1: Traverse from head to end

   Start at head (5), then move to 10, then 15, collecting values in order.

2. Step 2: Collect values in array during traversal

   Push each node's value into an array as you visit it, resulting in [5, 10, 15].

3. Final Answer:

   [5, 10, 15] -> Option C

4. Quick Check:

   Values collected in order visited [OK]

Hint: Collect values while traversing from head to null [OK]

Common Mistakes:

• Collecting values in reverse order
• Skipping nodes by wrong pointer update
• Returning array before traversal completes