Maximum Width of Binary Tree in DSA Go - Build from Scratch

Maximum Width of Binary Tree

📖 Scenario: You are working with a binary tree data structure that represents a family tree. You want to find the maximum width of this tree, which means the largest number of family members present at any single generation level.

🎯 Goal: Build a Go program that creates a binary tree, sets up a queue for level order traversal, calculates the maximum width of the tree, and prints the result.

📋 What You'll Learn

Create a binary tree with exactly 7 nodes with values 1 to 7 arranged as a complete binary tree

Create a queue to help with level order traversal

Implement the logic to find the maximum width of the binary tree

Print the maximum width as an integer

💡 Why This Matters

🌍 Real World

Binary trees are used in many applications like family trees, file systems, and decision trees. Knowing the maximum width helps understand the breadth of data at each level.

💼 Career

Understanding tree traversal and width calculation is important for software engineers working with hierarchical data, databases, and algorithms.

1

Create the binary tree nodes

Create a binary tree with a root node called root of type *TreeNode. The tree should have these exact nodes and structure:
1 as root, 2 as left child of root, 3 as right child of root, 4 as left child of node 2, 5 as right child of node 2, 6 as left child of node 3, and 7 as right child of node 3.

DSA Go

package main

import "fmt"

// TreeNode represents a node in the binary tree
type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func main() {
    // Create the binary tree nodes here
    // Your code here
}

Hint

Start by creating the root node with value 1. Then assign its left and right children with values 2 and 3 respectively. Continue assigning children for nodes 2 and 3 as described.

2

Create a queue for level order traversal

Create a slice of *TreeNode called queue and initialize it with the root node. This queue will help us visit nodes level by level.

DSA Go

package main

import "fmt"

// TreeNode represents a node in the binary tree
type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}
    root.Right.Left = &TreeNode{Val: 6}
    root.Right.Right = &TreeNode{Val: 7}

    // Create a queue slice with root node
    // Your code here
}

Hint

Initialize the queue slice with the root node inside curly braces.

3

Calculate the maximum width of the binary tree

Write code to calculate the maximum width of the binary tree using the queue. Use a variable maxWidth initialized to 0. Use a for loop that runs while len(queue) > 0. Inside the loop, get the current level size with levelSize := len(queue). Update maxWidth if levelSize is greater. Then, use a for loop from 0 to levelSize to dequeue nodes from queue and enqueue their children if they exist.

DSA Go

package main

import "fmt"

// TreeNode represents a node in the binary tree
type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}
    root.Right.Left = &TreeNode{Val: 6}
    root.Right.Right = &TreeNode{Val: 7}

    queue := []*TreeNode{root}

    // Calculate maxWidth using queue
    // Your code here
}

Hint

Use a while loop with len(queue) > 0. Inside, get the current level size. Update maxWidth if needed. Then dequeue each node of this level and enqueue their children.

4

Print the maximum width

Print the value of maxWidth using fmt.Println(maxWidth).

DSA Go

package main

import "fmt"

// TreeNode represents a node in the binary tree
type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}
    root.Right.Left = &TreeNode{Val: 6}
    root.Right.Right = &TreeNode{Val: 7}

    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)
            }
        }
    }

    // Print the maximum width
    // Your code here
}

Hint

Use fmt.Println(maxWidth) to print the maximum width.

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