Maximum Width of Binary Tree in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of width calculation for a simple binary tree

What is the maximum width of the binary tree after running the given code?

DSA Java

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

public class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        if (root == null) return 0;
        java.util.Queue<java.util.Map.Entry<TreeNode, Integer>> queue = new java.util.LinkedList<>();
        queue.offer(new java.util.AbstractMap.SimpleEntry<>(root, 0));
        int maxWidth = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            int minIndex = queue.peek().getValue();
            int first = 0, last = 0;
            for (int i = 0; i < size; i++) {
                java.util.Map.Entry<TreeNode, Integer> entry = queue.poll();
                TreeNode node = entry.getKey();
                int index = entry.getValue() - minIndex;
                if (i == 0) first = index;
                if (i == size - 1) last = index;
                if (node.left != null) queue.offer(new java.util.AbstractMap.SimpleEntry<>(node.left, 2 * index + 1));
                if (node.right != null) queue.offer(new java.util.AbstractMap.SimpleEntry<>(node.right, 2 * index + 2));
            }
            maxWidth = Math.max(maxWidth, last - first + 1);
        }
        return maxWidth;
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(3);
        root.right = new TreeNode(2);
        root.left.left = new TreeNode(5);
        root.left.right = new TreeNode(3);
        root.right.right = new TreeNode(9);
        Solution sol = new Solution();
        System.out.println(sol.widthOfBinaryTree(root));
    }
}

A 2

B 3

C 4

D 5

💡 Hint

Think about how the width is calculated including null positions between nodes.

The maximum width is calculated as the maximum distance between the leftmost and rightmost nodes at any level, counting nulls in between. Here, the bottom level has nodes at positions 0 (5), 1 (3), and 3 (9), so width is 3 - 0 + 1 = 4.

❓ Predict Output

intermediate

Output for a skewed binary tree

What is the maximum width of the following skewed binary tree after running the code?

DSA Java

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

public class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        if (root == null) return 0;
        java.util.Queue<java.util.Map.Entry<TreeNode, Integer>> queue = new java.util.LinkedList<>();
        queue.offer(new java.util.AbstractMap.SimpleEntry<>(root, 0));
        int maxWidth = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            int minIndex = queue.peek().getValue();
            int first = 0, last = 0;
            for (int i = 0; i < size; i++) {
                java.util.Map.Entry<TreeNode, Integer> entry = queue.poll();
                TreeNode node = entry.getKey();
                int index = entry.getValue() - minIndex;
                if (i == 0) first = index;
                if (i == size - 1) last = index;
                if (node.left != null) queue.offer(new java.util.AbstractMap.SimpleEntry<>(node.left, 2 * index + 1));
                if (node.right != null) queue.offer(new java.util.AbstractMap.SimpleEntry<>(node.right, 2 * index + 2));
            }
            maxWidth = Math.max(maxWidth, last - first + 1);
        }
        return maxWidth;
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.right = new TreeNode(2);
        root.right.right = new TreeNode(3);
        root.right.right.right = new TreeNode(4);
        Solution sol = new Solution();
        System.out.println(sol.widthOfBinaryTree(root));
    }
}

A 4

B 2

C 3

D 1

💡 Hint

In a skewed tree, width is usually 1 because nodes are aligned.

Each level has only one node, so the width at every level is 1. The maximum width is therefore 1.

🔧 Debug

advanced

Identify the error in the width calculation code

What error will the following code produce when run on a binary tree with many levels?

DSA Java

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

public class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        if (root == null) return 0;
        java.util.Queue<java.util.Map.Entry<TreeNode, Integer>> queue = new java.util.LinkedList<>();
        queue.offer(new java.util.AbstractMap.SimpleEntry<>(root, 0));
        int maxWidth = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            int minIndex = queue.peek().getValue();
            int first = 0, last = 0;
            for (int i = 0; i < size; i++) {
                java.util.Map.Entry<TreeNode, Integer> entry = queue.poll();
                TreeNode node = entry.getKey();
                int index = entry.getValue() - minIndex;
                if (i == 0) first = index;
                if (i == size - 1) last = index;
                if (node.left != null) queue.offer(new java.util.AbstractMap.SimpleEntry<>(node.left, 2 * index + 1));
                if (node.right != null) queue.offer(new java.util.AbstractMap.SimpleEntry<>(node.right, 2 * index + 2));
            }
            maxWidth = Math.max(maxWidth, last - first + 1);
        }
        return maxWidth;
    }
}

A NullPointerException when root is null

B Integer overflow causing wrong width calculation

C ArrayIndexOutOfBoundsException due to queue misuse

D StackOverflowError due to recursion

💡 Hint

Consider what happens to the index values at very deep levels.

The index values double at each level, so for very deep trees, the integer index can overflow causing incorrect calculations.

🧠 Conceptual

advanced

Understanding the indexing scheme in maximum width calculation

Why does the algorithm assign indices to nodes as 2 * index + 1 for left child and 2 * index + 2 for right child?

A To simulate the node positions as if the tree was a complete binary tree, preserving relative positions

B To assign unique IDs to nodes for hashing purposes

C To balance the tree by reordering nodes

D To reduce the memory usage of the queue

💡 Hint

Think about how the width is measured including gaps between nodes.

Assigning indices this way simulates the positions of nodes in a complete binary tree, allowing calculation of width including gaps caused by missing nodes.

❓ Predict Output

expert

Maximum width for a complex tree with missing nodes

What is the maximum width printed by the code for the following tree structure? Tree: - Root (1) - Left (2) - Right (5) - Right (3) - Right (4) - Left (6) - Right (7)

DSA Java

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

public class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        if (root == null) return 0;
        java.util.Queue<java.util.Map.Entry<TreeNode, Integer>> queue = new java.util.LinkedList<>();
        queue.offer(new java.util.AbstractMap.SimpleEntry<>(root, 0));
        int maxWidth = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            int minIndex = queue.peek().getValue();
            int first = 0, last = 0;
            for (int i = 0; i < size; i++) {
                java.util.Map.Entry<TreeNode, Integer> entry = queue.poll();
                TreeNode node = entry.getKey();
                int index = entry.getValue() - minIndex;
                if (i == 0) first = index;
                if (i == size - 1) last = index;
                if (node.left != null) queue.offer(new java.util.AbstractMap.SimpleEntry<>(node.left, 2 * index + 1));
                if (node.right != null) queue.offer(new java.util.AbstractMap.SimpleEntry<>(node.right, 2 * index + 2));
            }
            maxWidth = Math.max(maxWidth, last - first + 1);
        }
        return maxWidth;
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.right = new TreeNode(5);
        root.right.right = new TreeNode(4);
        root.right.right.left = new TreeNode(6);
        root.right.right.right = new TreeNode(7);
        Solution sol = new Solution();
        System.out.println(sol.widthOfBinaryTree(root));
    }
}

A 7

B 5

C 6

D 8

💡 Hint

Calculate indices level by level and find the widest gap including nulls.

The widest level is the last one with nodes at indices 9 (6) and 10 (7) and also nodes at earlier indices from previous levels, resulting in width 7.

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