Maximum Width of Binary Tree in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the queue with the root node and its index 0.

DSA Java

Queue<Pair<TreeNode, Integer>> queue = new LinkedList<>();
queue.[1](new Pair<>(root, 0));

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

Ainsert

Boffer

Cpush

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'push' which is for stacks, not queues.

Using 'insert' which is not a Queue method.

The 'offer' method adds an element to the queue safely and is preferred for queues in Java.

2fill in blank

medium

Complete the code to calculate the width of the current level.

DSA Java

int width = lastIndex - [1] + 1;

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AmaxWidth

BlevelSize

CcurrentIndex

DfirstIndex

Attempts:

3 left

💡 Hint

Common Mistakes

Using levelSize instead of firstIndex.

Forgetting to add 1 to the difference.

Width is calculated by subtracting the first index from the last index and adding 1.

3fill in blank

hard

Fix the error in the code to correctly assign the index to the left child node.

DSA Java

if (node.left != null) {
    queue.offer(new Pair<>(node.left, [1]));
}

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Aindex * 2

Bindex * 2 + 1

Cindex + 1

Dindex - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using index * 2 + 1 for left child which is for right child.

Using index + 1 which is incorrect for binary tree indexing.

Left child index is calculated as parent index * 2.

4fill in blank

hard

Fill both blanks to correctly assign the index to the right child node and update maxWidth.

DSA Java

if (node.right != null) {
    queue.offer(new Pair<>(node.right, [1]));
}
maxWidth = Math.[2](maxWidth, width);

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Aindex * 2 + 1

Bindex * 2

Cmax

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.min instead of Math.max.

Assigning wrong index to right child.

Right child index is parent index * 2 + 1. maxWidth is updated using Math.max to keep the largest width.

5fill in blank

hard

Fill all three blanks to complete the loop that processes each level of the tree.

DSA Java

while (!queue.isEmpty()) {
    int size = queue.[1]();
    int firstIndex = queue.peek().getValue();
    int lastIndex = 0;
    for (int i = 0; i < size; i++) {
        Pair<TreeNode, Integer> pair = queue.[2]();
        TreeNode node = pair.getKey();
        int index = pair.getValue() - firstIndex;
        if (node.left != null) queue.offer(new Pair<>(node.left, index * 2));
        if (node.right != null) queue.offer(new Pair<>(node.right, index * 2 + 1));
        if (i == size - 1) lastIndex = index;
    }
    maxWidth = Math.max(maxWidth, [3]);
}

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Asize

Bpoll

ClastIndex

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() instead of poll() inside the loop.

Using wrong variable for width calculation.

size is obtained by queue.size(), poll removes the head of the queue, and lastIndex is used to calculate width.

Practice

1. What does the "maximum width" of a binary tree represent?

easy

A. The largest number of nodes present at any single level of the tree

B. The height of the tree from root to the deepest leaf

C. The total number of nodes in the tree

D. The sum of values of all nodes at the widest level

Solution

1. Step 1: Understand the definition of width in a binary tree

   The width is the count of nodes at a particular level in the tree.

2. Step 2: Identify maximum width

   The maximum width is the largest such count among all levels.

3. Final Answer:

   The largest number of nodes present at any single level of the tree -> Option A

4. Quick Check:

   Maximum width = max nodes at any level [OK]

Hint: Width means count of nodes at one level, find max [OK]

Common Mistakes:

• Confusing width with tree height
• Thinking width is sum of node values
• Counting total nodes instead of per level

2. Which Java data structure is most suitable to track nodes level by level when finding maximum width of a binary tree?

easy

A. Stack

B. HashMap

C. Queue

D. ArrayList

Solution

1. Step 1: Identify traversal method for level order

   Level order traversal uses a queue to process nodes level by level.

2. Step 2: Choose data structure

   A queue allows nodes to be added and removed in FIFO order, perfect for level traversal.

3. Final Answer:

   Queue -> Option C

4. Quick Check:

   Level order traversal uses queue [OK]

Hint: Use queue for level order traversal to measure width [OK]

Common Mistakes:

• Using stack which is LIFO, not suitable for level order
• Choosing HashMap which doesn't preserve order
• Using ArrayList without queue behavior

3. What is the output of the following Java code snippet that calculates maximum width of a binary tree?

class TreeNode {
  int val;
  TreeNode left, right;
  TreeNode(int v) { val = v; }
}

// Tree structure:
//       1
//      / \
//     2   3
//    /     \
//   4       5

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

// Maximum width calculation
int maxWidth = 0;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
  int size = queue.size();
  maxWidth = Math.max(maxWidth, size);
  for (int i = 0; i < size; i++) {
    TreeNode node = queue.poll();
    if (node.left != null) queue.add(node.left);
    if (node.right != null) queue.add(node.right);
  }
}
System.out.println(maxWidth);

medium

A. 1

B. 2

C. 3

D. 4

Solution

1. Step 1: Analyze tree levels and node counts

   Level 1 has 1 node (1), level 2 has 2 nodes (2,3), level 3 has 2 nodes (4,5).

2. Step 2: Determine maximum width

   The maximum number of nodes at any level is 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Max width = 2 nodes at level 2 [OK]

Hint: Count nodes per level, max is answer [OK]

Common Mistakes:

• Counting total nodes instead of per level
• Assuming level 3 has 3 nodes incorrectly
• Ignoring null children in width calculation

4. Identify the bug in this Java code snippet that attempts to find the maximum width of a binary tree:

Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
int maxWidth = 0;
while (!queue.isEmpty()) {
  int size = queue.size();
  maxWidth = Math.max(maxWidth, size);
  for (int i = 0; i <= size; i++) {
    TreeNode node = queue.poll();
    if (node.left != null) queue.add(node.left);
    if (node.right != null) queue.add(node.right);
  }
}

medium

A. The loop condition should be i < size, not i <= size

B. The queue should be a Stack instead

C. maxWidth should be updated after the for loop

D. The root node is never added to the queue

Solution

1. Step 1: Check the for loop iteration count

   The loop runs from i = 0 to i <= size, which means it runs size+1 times, causing an extra poll.

2. Step 2: Correct the loop condition

   It should run exactly size times, so condition must be i < size.

3. Final Answer:

   The loop condition should be i < size, not i <= size -> Option A

4. Quick Check:

   Loop must run size times exactly [OK]

Hint: Use i < size for loops over queue size [OK]

Common Mistakes:

• Using <= instead of < in for loop
• Replacing queue with stack incorrectly
• Updating maxWidth at wrong place

5. Given a binary tree where some nodes are missing children, how can you modify the maximum width calculation to include null positions between nodes, so the width counts the distance between the leftmost and rightmost non-null nodes at each level? Choose the best approach.

hard

A. Calculate width by summing values of nodes at each level

B. Count only non-null nodes at each level ignoring positions

C. Use a stack to store nodes and count them

D. Assign an index to each node as if the tree was a complete binary tree, track min and max indices per level, width = max - min + 1

Solution

1. Step 1: Understand the problem with missing nodes

   Simple counting misses gaps caused by null children, so width is underestimated.

2. Step 2: Use indexing to simulate complete tree positions

   Assign indices to nodes as in a complete binary tree (root=0, left=2*index, right=2*index+1).

3. Step 3: Calculate width per level using min and max indices

   Width = max index - min index + 1, capturing gaps between nodes.

4. Final Answer:

   Assign an index to each node as if the tree was a complete binary tree, track min and max indices per level, width = max - min + 1 -> Option D

5. Quick Check:

   Indexing nodes captures gaps for accurate width [OK]

Hint: Use node indices to count gaps for true width [OK]

Common Mistakes:

• Ignoring null gaps between nodes
• Counting only non-null nodes without positions
• Using stack instead of queue for level traversal