Tree Traversal Level Order BFS in DSA Javascript - Build from Scratch

Tree Traversal Level Order BFS

📖 Scenario: You are working with a simple family tree. Each person has a name and may have children. You want to visit each family member level by level, starting from the oldest ancestor.

🎯 Goal: Build a program that performs a level order traversal (Breadth-First Search) on a tree structure and prints the names of family members level by level.

📋 What You'll Learn

Create a tree structure using JavaScript objects with name and children properties

Use a queue to help with level order traversal

Traverse the tree level by level and collect names in order

Print the names in the order they are visited

💡 Why This Matters

🌍 Real World

Level order traversal is used in many real-world applications like finding shortest paths in networks, organizing hierarchical data, and scheduling tasks.

💼 Career

Understanding tree traversals is important for software developers working with data structures, databases, and algorithms in interviews and real projects.

1

Create the family tree structure

Create a variable called familyTree that represents this tree:

- Root node with name "Grandpa"
- children array with two nodes:
- First child with name "Dad" and no children
- Second child with name "Uncle" and no children

DSA Javascript

// Create the familyTree object with the exact structure
// Your code here

Hint

Use nested objects with name and children properties. The children property is an array.

2

Create a queue for traversal

Create a variable called queue and set it to an array containing only the familyTree variable.

DSA Javascript

const familyTree = {
  name: "Grandpa",
  children: [
    { name: "Dad", children: [] },
    { name: "Uncle", children: [] }
  ]
};

// Create the queue array with familyTree inside
// Your code here

Hint

A queue can be an array where you add and remove elements from ends. Start with the root node inside.

3

Traverse the tree level by level

Create an empty array called result. Use a while loop that runs while queue.length is greater than 0. Inside the loop:
- Remove the first element from queue and call it current
- Add current.name to result
- Add all current.children to the end of queue using push(...current.children)

DSA Javascript

const familyTree = {
  name: "Grandpa",
  children: [
    { name: "Dad", children: [] },
    { name: "Uncle", children: [] }
  ]
};

const queue = [familyTree];

// Create an empty array result
// Use a while loop to traverse the tree
// Your code here

Hint

Use shift() to remove from front and push(...) to add children to the back of the queue.

4

Print the traversal result

Print the result array using console.log(result).

DSA Javascript

const familyTree = {
  name: "Grandpa",
  children: [
    { name: "Dad", children: [] },
    { name: "Uncle", children: [] }
  ]
};

const queue = [familyTree];
const result = [];

while (queue.length > 0) {
  const current = queue.shift();
  result.push(current.name);
  queue.push(...current.children);
}

// Print the result array
// Your code here

Hint

Use console.log(result) to see the names 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