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);

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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]();

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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);
}

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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]);
}

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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.