Tree Traversal Level Order BFS in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add the root node to the queue for level order traversal.

DSA Typescript

const queue: TreeNode[] = [];
queue.[1](root);

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

Aunshift

Bpop

Cshift

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push removes elements instead of adding.

Using shift or unshift changes the wrong end of the queue.

We use push to add the root node at the end of the queue.

2fill in blank

medium

Complete the code to remove the first node from the queue during BFS.

DSA Typescript

const currentNode = queue.[1]();

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

Ashift

Bunshift

Cpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop removes from the end, breaking BFS order.

Using push or unshift does not remove elements.

We use shift to remove the first element from the queue, following FIFO order.

3fill in blank

hard

Fix the error in adding child nodes to the queue only if they exist.

DSA Typescript

if (currentNode.left [1] null) {
  queue.push(currentNode.left);
}

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

A==

B!=

C!==

D===

Attempts:

3 left

💡 Hint

Common Mistakes

Using == or === can cause incorrect checks due to type coercion.

Using != is less strict and may allow undefined values.

We check if currentNode.left is not exactly null using !== before adding it.

4fill in blank

hard

Fill both blanks to collect node values in level order traversal.

DSA Typescript

const result: number[] = [];
while (queue.length > 0) {
  const currentNode = queue.shift()!;
  result.[1](currentNode.[2]);
}

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

Apush

Bpop

Cval

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop removes elements instead of adding.

Using incorrect property name like 'val' instead of 'value'.

We use push to add the current node's value to the result array.

5fill in blank

hard

Fill all three blanks to complete the level order BFS function returning node values.

DSA Typescript

function levelOrder(root: TreeNode | null): number[] {
  if (root === null) return [];
  const queue: TreeNode[] = [];
  const result: number[] = [];
  queue.[1](root);
  while (queue.length > 0) {
    const currentNode = queue.[2]();
    result.push(currentNode.[3]);
    if (currentNode.left !== null) queue.push(currentNode.left);
    if (currentNode.right !== null) queue.push(currentNode.right);
  }
  return result;
}

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

Apush

Bshift

Cvalue

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of shift breaks BFS order.

Accessing wrong property name for node value.

Adding root with shift instead of push.

The root is added with push, nodes removed with shift, and node values accessed via value.

Practice

1. What does the Level Order Traversal (BFS) of a binary tree do?

easy

A. Visits nodes in preorder (root, left, right)

B. Visits nodes level by level from top to bottom, left to right

C. Visits nodes in inorder (left, root, right)

D. Visits nodes in postorder (left, right, root)

Solution

1. Step 1: Understand Level Order Traversal

   Level Order Traversal visits nodes level by level starting from the root, moving left to right on each level.

2. Step 2: Compare with other traversals

   Preorder, inorder, and postorder are depth-first traversals, not level by level.

3. Final Answer:

   Visits nodes level by level from top to bottom, left to right -> Option B

4. Quick Check:

   Level Order = BFS = top to bottom, left to right [OK]

Hint: Level order means visit nodes by levels, not depth [OK]

Common Mistakes:

• Confusing level order with preorder traversal
• Thinking BFS visits nodes depth-first
• Mixing up traversal orders

2. Which of the following TypeScript code snippets correctly initializes a queue for BFS traversal?

easy

A. const queue: TreeNode[] = []; queue.push(root);

B. const queue = new Set(); queue.add(root);

C. const queue = new Map(); queue.set(0, root);

D. const queue = {}; queue.enqueue(root);

Solution

1. Step 1: Identify correct queue initialization

   In TypeScript, an array can be used as a queue with push and shift methods.

2. Step 2: Check each option

   const queue: TreeNode[] = []; queue.push(root); uses an array and push, which is correct. const queue = new Set(); queue.add(root); uses Set which is unordered. const queue = new Map(); queue.set(0, root); uses Map which is key-value, not queue. const queue = {}; queue.enqueue(root); uses an object with enqueue method which does not exist.

3. Final Answer:

   const queue: TreeNode[] = []; queue.push(root); -> Option A

4. Quick Check:

   Array with push/shift = queue [OK]

Hint: Use array with push and shift for queue in TypeScript [OK]

Common Mistakes:

• Using Set or Map instead of array for queue
• Calling non-existent enqueue method on object
• Not initializing queue before use

3. Given the binary tree below, what is the output of the level order traversal?

    1
   / \
  2   3
 /   / \
4   5   6

function levelOrder(root: TreeNode | null): number[] { if (!root) return []; const queue: TreeNode[] = [root]; const result: number[] = []; while (queue.length > 0) { const node = queue.shift()!; result.push(node.val); if (node.left) queue.push(node.left); if (node.right) queue.push(node.right); } return result; }

medium

A. [1, 2, 4, 3, 5, 6]

B. [1, 3, 2, 6, 5, 4]

C. [1, 2, 3, 4, 5, 6]

D. [4, 2, 5, 3, 6, 1]

Solution

1. Step 1: Trace the queue and result arrays

   Start with queue = [1], result = []. Remove 1, add to result: [1]. Add children 2 and 3 to queue: [2,3]. Remove 2, add to result: [1,2]. Add child 4: queue = [3,4]. Remove 3, add to result: [1,2,3]. Add children 5 and 6: queue = [4,5,6]. Remove 4, add to result: [1,2,3,4]. No children. Remove 5, add to result: [1,2,3,4,5]. No children. Remove 6, add to result: [1,2,3,4,5,6]. No children.

2. Step 2: Confirm final output

   The final result array is [1, 2, 3, 4, 5, 6].

3. Final Answer:

   [1, 2, 3, 4, 5, 6] -> Option C

4. Quick Check:

   Level order visits nodes level-wise: [1,2,3,4,5,6] [OK]

Hint: Follow queue order: push children left then right [OK]

Common Mistakes:

• Mixing left and right child order
• Confusing preorder with level order output
• Forgetting to add children to queue

4. The following TypeScript code for level order traversal has a bug. What is the error?

function levelOrder(root: TreeNode | null): number[] {
  const queue: TreeNode[] = [];
  const result: number[] = [];
  queue.push(root!);
  while (queue.length > 0) {
    const node = queue.shift();
    result.push(node.val);
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
  }
  return result;
}

medium

A. It pushes children after adding node to result

B. It uses shift() instead of pop() on the queue

C. It returns result before the loop ends

D. It does not check if root is null before pushing

Solution

1. Step 1: Analyze root null check

   The code pushes root! without checking if root is null. If root is null, queue contains null and node.val will cause runtime error.

2. Step 2: Verify other parts

   Using shift() is correct for queue. Adding children after node is correct. Return is after loop ends.

3. Final Answer:

   It does not check if root is null before pushing -> Option D

4. Quick Check:

   Always check root null before queue push [OK]

Hint: Check root null before queue push to avoid errors [OK]

Common Mistakes:

• Ignoring null root check
• Confusing shift() with pop() for queue
• Misplacing return statement

5. You want to print the level order traversal of a binary tree but separate each level on a new line. Which approach correctly achieves this in TypeScript?

function levelOrderByLevel(root: TreeNode | null): number[][] {
  if (!root) return [];
  const result: number[][] = [];
  const queue: TreeNode[] = [root];
  while (queue.length > 0) {
    const levelSize = queue.length;
    const currentLevel: number[] = [];
    for (let i = 0; i < levelSize; i++) {
      const node = queue.shift()!;
      currentLevel.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
    result.push(currentLevel);
  }
  return result;
}

hard

A. Use a loop for each level size, collect nodes, then push to result

B. Push all nodes to result as a flat array without levels

C. Use recursion to print nodes without queue

D. Add all children to queue before processing current level

Solution

1. Step 1: Understand level separation logic

   Using levelSize to track nodes at current level allows processing nodes level by level.

2. Step 2: Check code correctness

   Loop runs levelSize times, collects node values in currentLevel, adds children to queue for next level, then pushes currentLevel to result.

3. Final Answer:

   Use a loop for each level size, collect nodes, then push to result -> Option A

4. Quick Check:

   Track level size, process nodes per level [OK]

Hint: Use queue length to separate levels in BFS [OK]

Common Mistakes:

• Not separating levels, returning flat array
• Using recursion instead of queue for BFS
• Adding children before processing current level nodes