Maximum Width of Binary Tree in DSA Go - 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 (maximum position - minimum position + 1) of nodes at that level.

intermediate

How can we keep track of node positions to calculate the maximum width in a binary tree?

We assign a position index to each node starting from 0 at the root. For any node at position p, its left child is at 2*p + 1 and right child at 2*p + 2. This helps measure the width by subtracting the minimum position from the maximum position at each level.

intermediate

Why do we subtract the minimum position from the maximum position at each level when calculating width?

Subtracting the minimum position from the maximum position gives the number of positions spanned by nodes at that level. Adding 1 gives the total count of positions including gaps, which represents the width.

beginner

What data structure is commonly used to traverse the tree level by level for width calculation?

A queue is used to perform a level order traversal (breadth-first search) to visit nodes level by level and track their positions.

beginner

In Go, how do you represent a binary tree node for this problem?

A binary tree node is represented as a struct with an integer value and pointers to left and right child nodes, for example: type TreeNode struct { Val int Left *TreeNode Right *TreeNode }

What is the position of the right child if the parent node's position is 3?

A 5

B 8

C 6

D 7

Right child position = 2 * parent_position + 2 = 2 * 3 + 2 = 8.

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

A Level order traversal

B Inorder traversal

C Postorder traversal

D Preorder traversal

Level order traversal visits nodes level by level, which is needed to measure width at each level.

If a level has nodes at positions 4, 5, and 7, what is the width of that level?

A 2

B 4

C 3

D 5

Width = max_position - min_position + 1 = 7 - 4 + 1 = 4.

What initial position index is assigned to the root node?

A 0

B 1

C -1

D None

The root node is assigned position 0 to start indexing from zero.

Why do we use a queue in the maximum width calculation algorithm?

A To store nodes in sorted order

B To reverse the tree

C To traverse nodes depth-first

D To traverse nodes level by level

A queue helps traverse nodes level by level (breadth-first), which is essential for width calculation.

Explain how to calculate the maximum width of a binary tree using position indexing.

Think about numbering nodes and measuring gaps at each level.

Describe the role of a queue in finding the maximum width of a binary tree.

Consider how to visit nodes level by level.

Practice

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

easy

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

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

C. The total number of nodes in the tree

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

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 D

4. Quick Check:

   Maximum width = max nodes at any level [OK]

Hint: Width means count of nodes at one level, not height or sum [OK]

Common Mistakes:

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

2. Which Go data structure is best suited to track nodes level by level when finding maximum width of a binary tree?

easy

A. Queue

B. Stack

C. Map

D. Slice

Solution

1. Step 1: Identify traversal method for level order

   Level order traversal requires processing nodes level by level, which is done using a queue.

2. Step 2: Choose data structure

   A queue allows nodes to be added at the end and removed from the front, perfect for breadth-first traversal.

3. Final Answer:

   Queue -> Option A

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
• Using map which doesn't preserve order
• Using slice without queue operations

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

type TreeNode struct {
    Val int
    Left, Right *TreeNode
}

func maxWidth(root *TreeNode) int {
    if root == nil {
        return 0
    }
    queue := []*TreeNode{root}
    maxWidth := 0
    for len(queue) > 0 {
        levelSize := len(queue)
        if levelSize > maxWidth {
            maxWidth = levelSize
        }
        for i := 0; i < levelSize; i++ {
            node := queue[0]
            queue = queue[1:]
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
    }
    return maxWidth
}

root := &TreeNode{Val: 1}
root.Left = &TreeNode{Val: 2}
root.Right = &TreeNode{Val: 3}
root.Left.Left = &TreeNode{Val: 4}
root.Right.Right = &TreeNode{Val: 5}

fmt.Println(maxWidth(root))

medium

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Analyze tree structure and levels

   Level 1: Node 1 (1 node), Level 2: Nodes 2 and 3 (2 nodes), Level 3: Nodes 4 and 5 (2 nodes)

2. Step 2: Calculate maximum width

   Maximum width is the largest number of nodes at any level, which is 2.

3. Step 3: Re-examine the tree structure carefully

   Actually, at Level 3, nodes are 4 and 5 (2 nodes), but at Level 2, nodes are 2 and 3 (2 nodes), so max width is 2. However, the code counts the queue size before processing children, so at Level 1 queue size is 1, Level 2 queue size is 2, Level 3 queue size is 2. The maximum width is 2.

4. Step 4: Check if any level has 3 nodes

   No level has 3 nodes, so 3 (3) is incorrect.

5. Final Answer:

   2 -> Option B

6. Quick Check:

   Max nodes at any level = 2 [OK]

Hint: Count nodes per level carefully to find max width [OK]

Common Mistakes:

• Counting total nodes instead of per level
• Miscounting nodes at last level
• Assuming root level has more nodes

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

func maxWidth(root *TreeNode) int {
    if root == nil {
        return 0
    }
    queue := []*TreeNode{root}
    maxWidth := 0
    for len(queue) > 0 {
        levelSize := len(queue)
        if levelSize > maxWidth {
            maxWidth = levelSize
        }
        for i := 0; i <= levelSize; i++ {
            node := queue[0]
            queue = queue[1:]
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
    }
    return maxWidth
}

medium

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

B. The queue should be initialized empty, not with root

C. The maxWidth should be updated after the inner loop

D. The function should return maxWidth + 1

Solution

1. Step 1: Check loop boundaries for processing nodes

   The inner loop runs from i = 0 to i <= levelSize, which means it runs levelSize + 1 times, causing out-of-range error.

2. Step 2: Correct loop condition

   The loop should run exactly levelSize times, so condition must be i < levelSize.

3. Final Answer:

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

4. Quick Check:

   Loop runs one extra time causing error [OK]

Hint: Use < for loops when iterating fixed count [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Initializing queue incorrectly
• Updating maxWidth at wrong time

5. Given a binary tree where some nodes are missing children, how can you modify the maximum width calculation to include the 'null' positions between nodes, effectively measuring the width as the distance between the leftmost and rightmost non-null nodes at each level? Choose the best approach to implement this in Go.

hard

A. Use recursion to count nodes and ignore null gaps

B. Count only non-null nodes per level ignoring positions

C. Use a queue storing pairs of (node, index) where index represents position, and calculate width as max index difference per level

D. Calculate width as total nodes in tree divided by height

Solution

1. Step 1: Understand the problem of null gaps

   Normal width counts nodes only, but to include gaps, we track positions of nodes as if tree was complete.

2. Step 2: Use index tracking with queue

   Store each node with an index representing its position. For each level, width = last index - first index + 1.

3. Step 3: Implement with queue of (node, index) pairs

   This approach allows measuring width including null gaps between nodes.

4. Final Answer:

   Use a queue storing pairs of (node, index) where index represents position, and calculate width as max index difference per level -> Option C

5. Quick Check:

   Index tracking captures null gaps in width [OK]

Hint: Track node positions with indices to include gaps [OK]

Common Mistakes:

• Ignoring null gaps by counting only nodes
• Using recursion without position tracking
• Calculating width as total nodes divided by height