Tree Traversal Level Order BFS in DSA Javascript - Execution Trace

Concept Flow - Tree Traversal Level Order BFS

Start at root node

↓

Add root to queue

↓

While queue not empty

↓

Remove front node from queue

↓

Visit node (process value)

↓

Add left child to queue if exists

↓

Add right child to queue if exists

↓

Repeat loop

↓

Queue empty -> traversal done

Start from the root, use a queue to visit nodes level by level, adding children to the queue as you go.

Execution Sample

DSA Javascript

function levelOrder(root) {
  const result = [];
  const queue = [];
  if (root) 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;
}

This code visits each node of a binary tree level by level and collects their values in order.

Execution Table

Step	Operation	Nodes in Queue (values)	Node Visited	Queue Changes	Visual State (Tree nodes visited)
1	Start with root	[1]	None	Add root (1) to queue	[]
2	Dequeue node	[1]	1	Remove 1 from queue	[1]
3	Visit node 1	[]	1	Add children 2, 3 to queue	[1]
4	Dequeue node	[2, 3]	2	Remove 2 from queue	[1, 2]
5	Visit node 2	[3]	2	Add children 4, 5 to queue	[1, 2]
6	Dequeue node	[3, 4, 5]	3	Remove 3 from queue	[1, 2, 3]
7	Visit node 3	[4, 5]	3	Add children 6, 7 to queue	[1, 2, 3]
8	Dequeue node	[4, 5, 6, 7]	4	Remove 4 from queue	[1, 2, 3, 4]
9	Visit node 4	[5, 6, 7]	4	No children to add	[1, 2, 3, 4]
10	Dequeue node	[5, 6, 7]	5	Remove 5 from queue	[1, 2, 3, 4, 5]
11	Visit node 5	[6, 7]	5	No children to add	[1, 2, 3, 4, 5]
12	Dequeue node	[6, 7]	6	Remove 6 from queue	[1, 2, 3, 4, 5, 6]
13	Visit node 6	[7]	6	No children to add	[1, 2, 3, 4, 5, 6]
14	Dequeue node	[7]	7	Remove 7 from queue	[1, 2, 3, 4, 5, 6, 7]
15	Visit node 7	[]	7	No children to add	[1, 2, 3, 4, 5, 6, 7]
16	Queue empty	[]	None	Traversal complete	[1, 2, 3, 4, 5, 6, 7]

💡 Queue is empty, all nodes visited in level order

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 5	After Step 7	After Step 9	After Step 11	After Step 13	After Step 15	Final
queue	[]	[1]	[2, 3]	[3, 4, 5]	[4, 5, 6, 7]	[5, 6, 7]	[6, 7]	[7]	[]	[]
result	[]	[]	[1]	[1, 2]	[1, 2, 3]	[1, 2, 3, 4]	[1, 2, 3, 4, 5]	[1, 2, 3, 4, 5, 6]	[1, 2, 3, 4, 5, 6, 7]	[1, 2, 3, 4, 5, 6, 7]
node	null	1	2	3	4	5	6	7	null	null

Key Moments - 3 Insights

Why do we add children to the queue only after visiting the current node?

Because in the execution_table rows 3, 5, 7, etc., we first remove the node from the queue, then visit it, and only then add its children. This ensures nodes are visited level by level.

What happens if the tree is empty (root is null)?

The queue never gets any node added (see variable_tracker start state), so the while loop never runs and the traversal ends immediately with an empty result.

Why do we use a queue and not a stack for level order traversal?

A queue processes nodes in the order they were added (FIFO), which matches visiting nodes level by level. Using a stack (LIFO) would visit nodes depth-first, not level order.

Visual Quiz - 3 Questions

Look at the execution_table at Step 7, what nodes are currently in the queue?

A [3, 4, 5]

B [5, 6, 7]

C [4, 5, 6, 7]

D [2, 3]

💡 Hint

Check the 'Nodes in Queue (values)' column at Step 7 in the execution_table.

At which step does the queue become empty, ending the traversal?

A Step 15

B Step 16

C Step 14

D Step 13

💡 Hint

Look for the step where 'Nodes in Queue' is empty and the exit_note says traversal complete.

If node 3 had no children, how would the queue change after Step 7?

A [4, 5]

B [4, 5, 6, 7]

C [4, 5, 6]

D [5, 6, 7]

💡 Hint

Refer to Step 7 where children of node 3 are added; removing them means fewer nodes added.

Concept Snapshot

Level Order Traversal (BFS) visits tree nodes level by level.
Use a queue to hold nodes to visit.
Start by adding root to queue.
While queue not empty: dequeue node, visit it, enqueue children.
Traversal ends when queue is empty.
Result is nodes visited in level order.

Full Transcript

Level Order Traversal uses a queue to visit nodes of a binary tree one level at a time. We start by adding the root node to the queue. Then, while the queue is not empty, we remove the front node, visit it by recording its value, and add its left and right children to the queue if they exist. This process repeats until the queue is empty, meaning all nodes have been visited in level order. The execution table shows each step, the queue contents, the node visited, and the visual state of visited nodes. The variable tracker follows the queue, result list, and current node through the steps. Key moments clarify why children are added after visiting, what happens if the tree is empty, and why a queue is used instead of a stack. The visual quiz tests understanding of queue contents at specific steps and effects of missing children. The concept snapshot summarizes the traversal process simply and clearly.

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