Tree Traversal Level Order BFS in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this level order traversal?

Given the binary tree below, what is the printed output of the level order traversal function?

DSA Javascript

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

function levelOrder(root) {
  if (!root) return [];
  const queue = [root];
  const result = [];
  while (queue.length > 0) {
    const node = queue.shift();
    result.push(node.value);
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
  }
  return result;
}

// Tree structure:
//       10
//      /  \
//     5    15
//    / \     \
//   3   7     18

const root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);
root.left.left = new Node(3);
root.left.right = new Node(7);
root.right.right = new Node(18);

console.log(levelOrder(root));

A [10, 5, 3, 7, 15, 18]

B [10, 5, 15, 3, 7, 18]

C [10, 15, 5, 18, 7, 3]

D [3, 7, 5, 18, 15, 10]

💡 Hint

Think about visiting nodes level by level from left to right.

Level order traversal visits nodes level by level starting from the root, then left to right at each level. So the order is 10 (root), then 5 and 15 (level 2), then 3, 7, and 18 (level 3).

❓ Predict Output

intermediate

What is the output of this level order traversal with null nodes?

What will be the output of the level order traversal function when the tree has some missing children?

DSA Javascript

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

function levelOrder(root) {
  if (!root) return [];
  const queue = [root];
  const result = [];
  while (queue.length > 0) {
    const node = queue.shift();
    result.push(node.value);
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
  }
  return result;
}

// Tree structure:
//       1
//      / \
//     2   null
//    / \
// null 3

const root = new Node(1);
root.left = new Node(2);
root.left.right = new Node(3);

console.log(levelOrder(root));

A [1, 2, 3]

B [1, 2]

C [1, 3, 2]

D [2, 1, 3]

💡 Hint

Null children are skipped and not added to the queue.

The traversal visits 1 first, then 2 (left child), then 3 (right child of 2). Missing children are ignored.

🧠 Conceptual

advanced

Which data structure is essential for implementing level order traversal?

To perform a level order traversal (BFS) on a binary tree, which data structure is most suitable to keep track of nodes to visit next?

A Queue

B Set

C Hash Map

D Stack

💡 Hint

Think about visiting nodes in the order they appear level by level.

A queue follows First-In-First-Out order, which matches the level order traversal visiting nodes level by level.

🔧 Debug

advanced

What error does this level order traversal code produce?

Identify the error when running this level order traversal code snippet.

DSA Javascript

function levelOrder(root) {
  if (!root) return [];
  const queue = [];
  const result = [];
  queue.push(root);
  while (queue.length > 0) {
    const node = queue.pop();
    result.push(node.value);
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
  }
  return result;
}

A SyntaxError due to missing semicolon

B Infinite loop because queue never empties

C TypeError because queue is undefined

D The traversal order is reversed, not level order

💡 Hint

Check how nodes are removed from the queue.

Using pop() removes the last element, making the traversal depth-first (stack behavior), reversing the order.

🚀 Application

expert

How many nodes are visited in this level order traversal?

Given the following tree, how many nodes will the level order traversal visit?

DSA Javascript

// Tree structure:
//       8
//      / \
//     4   12
//    /   /  \
//   2   10  14
//        
//       11

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

const root = new Node(8);
root.left = new Node(4);
root.right = new Node(12);
root.left.left = new Node(2);
root.right.left = new Node(10);
root.right.right = new Node(14);
root.right.left.right = new Node(11);

function levelOrder(root) {
  if (!root) return [];
  const queue = [root];
  const result = [];
  while (queue.length > 0) {
    const node = queue.shift();
    result.push(node.value);
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
  }
  return result;
}

const visitedNodes = levelOrder(root).length;
console.log(visitedNodes);

A 8

B 6

C 7

D 5

💡 Hint

Count all nodes including the root and all children.

The tree has nodes: 8,4,12,2,10,14,11 totaling 7 nodes visited in level order.

Practice

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

easy

A. Visits nodes in left subtree, then root, then right subtree

B. Visits nodes in right subtree, then root, then left subtree

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

D. Visits nodes randomly without any order

Solution

1. Step 1: Understand Level Order Traversal

   Level order traversal visits nodes starting from the root, then all nodes at the next level, and so on, moving left to right within each level.

2. Step 2: Compare options with definition

   Visits nodes level by level from top to bottom, left to right matches this definition exactly, while others describe different traversals or incorrect behavior.

3. Final Answer:

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

4. Quick Check:

   Level Order = Top to bottom, left to right [OK]

Hint: Level order means visit nodes level-wise, left to right [OK]

Common Mistakes:

• Confusing level order with inorder or preorder traversal
• Thinking BFS visits nodes randomly
• Mixing left and right subtree orders

2. Which of the following is the correct way to initialize a queue for BFS in JavaScript?

easy

A. const queue = null

B. const queue = {}

C. const queue = 0

D. const queue = []

Solution

1. Step 1: Identify data structure for BFS queue

   BFS uses a queue to hold nodes to visit. In JavaScript, an array can be used as a queue with push and shift methods.

2. Step 2: Check options for correct queue initialization

   const queue = [] initializes an empty array, suitable for queue operations. Others are incorrect types for queue usage.

3. Final Answer:

   const queue = [] -> Option D

4. Quick Check:

   Queue = array for BFS [OK]

Hint: Use an empty array [] as queue for BFS in JavaScript [OK]

Common Mistakes:

• Using object {} instead of array for queue
• Initializing queue as null or number
• Not knowing how to enqueue/dequeue in JS

3. What is the output of the following code snippet?

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);

function levelOrder(root) {
  const result = [];
  const queue = [root];
  while (queue.length) {
    const node = queue.shift();
    result.push(node.val);
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
  }
  return result;
}

console.log(levelOrder(root));

medium

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

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

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

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

Solution

1. Step 1: Trace the BFS traversal on the tree

   Start with root (1), then enqueue left (2) and right (3). Next dequeue 2, enqueue its children 4 and 5. Then dequeue 3 (no children). Then dequeue 4 and 5.

2. Step 2: Collect visited nodes in order

   Visited order is 1, 2, 3, 4, 5 as nodes are processed level by level, left to right.

3. Final Answer:

   [1, 2, 3, 4, 5] -> Option A

4. Quick Check:

   Level order output = [1, 2, 3, 4, 5] [OK]

Hint: Follow queue order: root, left child, right child [OK]

Common Mistakes:

• Mixing preorder or inorder output with BFS
• Forgetting to enqueue children in correct order
• Confusing shift() with pop() in queue

4. Find the error in this BFS code snippet for a binary tree:

function bfs(root) {
  const queue = [];
  queue.push(root);
  while (queue.length > 0) {
    const node = queue.pop();
    console.log(node.val);
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
  }
}

medium

A. Not checking if root is null before pushing

B. Using pop() instead of shift() causes wrong traversal order

C. Missing return statement

D. Using console.log instead of return

Solution

1. Step 1: Identify queue operation used

   The code uses pop() to remove nodes from the queue, which removes from the end (stack behavior).

2. Step 2: Understand BFS queue behavior

   BFS requires removing nodes from the front (shift()), not the end. Using pop() reverses order, causing DFS-like traversal.

3. Final Answer:

   Using pop() instead of shift() causes wrong traversal order -> Option B

4. Quick Check:

   BFS needs shift(), not pop() [OK]

Hint: Use shift() to dequeue in BFS, not pop() [OK]

Common Mistakes:

• Confusing pop() with shift() in queue
• Ignoring null root check (less critical here)
• Expecting return instead of console.log (not error)

5. Given a binary tree, how can you modify level order traversal to return nodes level by level as nested arrays?
Example tree:

    1
   / \
  2   3
 / \
4   5

hard

A. Use a queue and track the number of nodes at each level to group values

B. Use recursion to visit nodes in preorder and group by depth

C. Use a stack to reverse the order of nodes after traversal

D. Traverse left subtree fully before right subtree and collect nodes

Solution

1. Step 1: Understand grouping nodes by level

   To return nodes level by level, we must know how many nodes are at each level during traversal.

2. Step 2: Use queue size to separate levels

   At each iteration, record queue length (nodes at current level), dequeue exactly that many nodes, collect their values, then enqueue their children for next level.

3. Step 3: Compare options

   Use a queue and track the number of nodes at each level to group values describes this approach correctly. Others describe preorder recursion, stack reversal, or incorrect traversal orders.

4. Final Answer:

   Use a queue and track the number of nodes at each level to group values -> Option A

5. Quick Check:

   Queue + level size tracking = nested level arrays [OK]

Hint: Track queue length each loop to separate levels [OK]

Common Mistakes:

• Trying to use recursion without level info
• Using stack which reverses order incorrectly
• Not grouping nodes, returning flat list instead