Tree Traversal Level Order BFS in DSA Java - 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 Java

Queue<TreeNode> queue = new LinkedList<>();
queue.[1](root);

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

Apeek

Badd

Cremove

Dpoll

Attempts:

3 left

💡 Hint

Common Mistakes

Using remove() instead of add() to insert the root.

Using peek() which only views the element without adding.

Using poll() which removes the element instead of adding.

We use queue.add(root) to start the level order traversal by adding the root node to the queue.

2fill in blank

medium

Complete the code to get the current node from the queue during traversal.

DSA Java

TreeNode current = queue.[1]();

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

Apoll

Badd

Coffer

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using add() which inserts instead of removing.

Using offer() which inserts instead of removing.

Using peek() which only views without removing.

queue.poll() removes and returns the head of the queue, which is the current node to process.

3fill in blank

hard

Fix the error in adding the left child node to the queue only if it exists.

DSA Java

if (current.[1] != null) {
    queue.add(current.left);
}

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

Aright

Bparent

Cchild

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Checking current.right instead of current.left.

Using a non-existent property like 'parent' or 'child'.

We check if current.left is not null before adding it to the queue to avoid errors.

4fill in blank

hard

Fill both blanks to complete the loop that processes nodes until the queue is empty.

DSA Java

while (!queue.[1]()) {
    TreeNode current = queue.[2]();
    // process current
}

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

AisEmpty

Bpoll

Cpeek

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() instead of poll() to remove nodes.

Using add() inside the while condition which is incorrect.

The loop continues while the queue is not empty (isEmpty() returns false). Inside, poll() removes the current node.

5fill in blank

hard

Fill all three blanks to build a map of node values to their level during BFS traversal.

DSA Java

Map<TreeNode, Integer> levels = new HashMap<>();
levels.put(root, [1]);
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);

while (!queue.isEmpty()) {
    TreeNode current = queue.poll();
    int currentLevel = levels.get(current);
    if (current.left != null) {
        queue.add(current.left);
        levels.put(current.left, currentLevel [2] [3]);
    }
    if (current.right != null) {
        queue.add(current.right);
        levels.put(current.right, currentLevel [2] [3]);
    }
}

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

A0

B+

C1

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Starting root level at 1 instead of 0.

Subtracting 1 instead of adding 1 for child levels.

Using multiplication or other operators instead of addition.

The root is at level 0. Each child is at parent's level + 1, so we add 1 to currentLevel.

Practice

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

easy

A. Visits nodes in random order

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

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

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

Solution

1. Step 1: Understand Level Order Traversal

   Level order traversal visits nodes level by level starting from the root, moving left to right at each level.

2. Step 2: Compare with other traversals

   Other traversals like inorder or postorder do not visit nodes level-wise but follow subtree orders.

3. Final Answer:

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

4. Quick Check:

   Level Order = BFS = top-down, left-right [OK]

Hint: Level order means visit nodes by levels, left to right [OK]

Common Mistakes:

• Confusing level order with inorder or preorder traversal
• Thinking traversal is random
• Mixing left and right subtree orders

2. Which Java code snippet correctly initializes a queue for level order traversal of a binary tree?

easy

A. Queue<TreeNode> queue = new ArrayList<>();

B. Queue<TreeNode> queue = new LinkedList<>();

C. List<TreeNode> queue = new LinkedList<>();

D. Stack<TreeNode> queue = new Stack<>();

Solution

1. Step 1: Identify correct queue initialization

   Java's Queue interface is implemented by LinkedList, so Queue<TreeNode> queue = new LinkedList<>() is correct.

2. Step 2: Check other options

   ArrayList does not implement Queue, List is not a Queue, and Stack is LIFO, not FIFO needed for BFS.

3. Final Answer:

   Queue<TreeNode> queue = new LinkedList<>(); -> Option B

4. Quick Check:

   Queue for BFS = LinkedList implementation [OK]

Hint: Use LinkedList to implement Queue for BFS in Java [OK]

Common Mistakes:

• Using ArrayList instead of Queue
• Using Stack which is LIFO, not FIFO
• Declaring List instead of Queue

3. What is the output of the following Java code performing level order traversal on this tree?

    1
   / \
  2   3
 / \
4   5

// Code snippet:
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
  TreeNode node = queue.poll();
  System.out.print(node.val + " ");
  if (node.left != null) queue.add(node.left);
  if (node.right != null) queue.add(node.right);
}

medium

A. 1 2 3 4 5

B. 1 3 2 5 4

C. 4 5 2 3 1

D. 2 4 5 1 3

Solution

1. Step 1: Trace the queue operations

   Start with root (1). Print 1, add children 2 and 3. Then print 2, add 4 and 5. Then print 3 (no children). Then print 4 and 5.

2. Step 2: Write output sequence

   Output is: 1 2 3 4 5

3. Final Answer:

   1 2 3 4 5 -> Option A

4. Quick Check:

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

Hint: Print nodes as you dequeue in BFS order [OK]

Common Mistakes:

• Mixing preorder or inorder output with level order
• Adding children in wrong order
• Printing before adding children

4. Identify the bug in this Java code snippet for level order traversal:

Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
  TreeNode node = queue.peek();
  System.out.print(node.val + " ");
  if (node.left != null) queue.add(node.left);
  if (node.right != null) queue.add(node.right);
}

medium

A. Using peek() instead of poll() causes infinite loop

B. Not checking if root is null

C. Adding children after printing node value

D. Queue should be Stack for BFS

Solution

1. Step 1: Understand peek() vs poll()

   peek() returns the front element but does not remove it, so the queue never empties.

2. Step 2: Effect on loop

   Since queue is never emptied, while loop runs infinitely printing the same node repeatedly.

3. Final Answer:

   Using peek() instead of poll() causes infinite loop -> Option A

4. Quick Check:

   poll() removes front, peek() does not [OK]

Hint: Use poll() to remove nodes, not peek() [OK]

Common Mistakes:

• Confusing peek() and poll()
• Ignoring null root check
• Using Stack instead of Queue

5. Given a binary tree, how can you modify level order traversal to print nodes level by level on separate lines in Java?

hard

A. Print nodes as they are dequeued without tracking levels

B. Use recursion to print nodes in preorder with newlines

C. Use a stack to reverse the order and print levels bottom-up

D. Use a queue and track the number of nodes at each level, printing a newline after each level

Solution

1. Step 1: Track nodes per level using queue size

   At each iteration, get the current queue size which equals nodes at that level. Process exactly that many nodes, then print newline.

2. Step 2: Print nodes level-wise

   This ensures nodes of one level print on one line, then move to next line for next level.

3. Final Answer:

   Use a queue and track the number of nodes at each level, printing a newline after each level -> Option D

4. Quick Check:

   Queue size = nodes per level for line breaks [OK]

Hint: Use queue size to separate levels during BFS [OK]

Common Mistakes:

• Using recursion which is not level order
• Using stack which reverses order
• Not tracking level size, so no line breaks