Maximum Width of Binary Tree in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the "maximum width" of a binary tree mean?

The maximum width of a binary tree is the maximum over all levels of (rightmost node's index - leftmost node's index + 1).

beginner

How can we find the width of each level in a binary tree?

We can find the width of each level during a level-order traversal by tracking the minimum and maximum indices of nodes at that level and computing (max_index - min_index + 1).

intermediate

Why do we assign indices to nodes when calculating maximum width of a binary tree?

Assigning indices to nodes helps to account for missing nodes between the leftmost and rightmost nodes at a level, so we can calculate the width including gaps.

beginner

What data structure is commonly used to perform level-order traversal for maximum width calculation?

A queue is commonly used to perform level-order traversal, allowing us to process nodes level by level.

intermediate

In Java, what is the purpose of using a Pair or custom class to store node and index during traversal?

It stores both the tree node and its assigned index together, so we can track the position of nodes to calculate width correctly.

What does the maximum width of a binary tree represent?

A The height of the tree

B The total number of nodes

C The maximum (rightmost index - leftmost index + 1) over all levels

D The number of leaf nodes

Maximum width is the largest (rightmost index - leftmost index + 1) among all levels.

Which traversal method is best suited to find the maximum width of a binary tree?

A Pre-order traversal

B Level-order traversal

C In-order traversal

D Post-order traversal

Level-order traversal visits nodes level by level, making it easy to count nodes per level.

Why do we assign indices to nodes during traversal for maximum width calculation?

A To track node positions including gaps

B To sort nodes by value

C To count total nodes

D To find the height of the tree

Indices help include gaps between nodes when calculating width.

What data structure is typically used to hold nodes during level-order traversal?

A Queue

B Stack

C Array

D HashMap

Queue allows processing nodes in FIFO order, perfect for level-order traversal.

In Java, what does a Pair typically represent in this problem?

A A node and its parent

B A node and its value

C Two nodes connected

D A tree node and its index

The pair stores the node and its assigned index to calculate width.

Explain how to calculate the maximum width of a binary tree using level-order traversal and node indexing.

Think about how to count nodes including gaps at each level.

Describe why simple counting of nodes at each level might not give the correct maximum width in a binary tree.

Consider a level with nodes only at the edges but missing in the middle.

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