Maximum Width of Binary Tree in DSA Go - Execution Trace

Concept Flow - Maximum Width of Binary Tree

Start at root node

↓

Initialize queue with root and index 0

↓

While queue not empty

↓

For each level: get first and last node indices

↓

Calculate width = last_index - first_index + 1

↓

Update max width if current width is larger

↓

Add children to queue with updated indices

↓

Repeat for next level

↓

Return max width found

We start from the root, use a queue to track nodes level by level with indices, calculate width per level, update max width, and continue until all levels are processed.

Execution Sample

DSA Go

func widthOfBinaryTree(root *TreeNode) int {
    maxWidth := 0
    queue := []struct{node *TreeNode; index int}{{root, 0}}
    for len(queue) > 0 {
        levelLen := len(queue)
        startIndex := queue[0].index
        endIndex := queue[levelLen-1].index
        width := endIndex - startIndex + 1
        if width > maxWidth {
            maxWidth = width
        }
        for i := 0; i < levelLen; i++ {
            curr := queue[0]
            queue = queue[1:]
            if curr.node.Left != nil {
                queue = append(queue, struct{node *TreeNode; index int}{curr.node.Left, 2 * curr.index + 1})
            }
            if curr.node.Right != nil {
                queue = append(queue, struct{node *TreeNode; index int}{curr.node.Right, 2 * curr.index + 2})
            }
        }
    }
    return maxWidth
}

This code finds the maximum width of a binary tree by tracking node indices at each level using a queue.

Execution Table

Step	Operation	Nodes in Queue (node val:index)	Level Length	Start Index	End Index	Width Calculated	Max Width	Visual Tree State
1	Initialize queue with root	[1:0]	1	0	0	1	1	Level 0: 1(0)
2	Process level 0	[1:0]	1	0	0	1	1	Level 0 processed, enqueue children 2:1, 3:2
3	Process level 1	[2:1, 3:2]	2	1	2	2	2	Level 1 processed, enqueue children 4:3, 5:4, 6:5
4	Process level 2	[4:3, 5:4, 6:5]	3	3	5	3	3	Level 2 processed, enqueue child 7:9
5	Process level 3	[7:9]	1	9	9	1	3	Level 3 processed, no children
6	Queue empty, end	[]	0	-	-	-	3	Traversal complete, max width = 3

💡 Queue is empty, all levels processed, maximum width found is 3

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	Final
maxWidth	0	1	2	3	3	3	3
queue	[(1,0)]	[(2,1),(3,2)]	[(4,3),(5,4),(6,5)]	[(7,9)]	[]	[]	[]

Key Moments - 3 Insights

Why do we assign indices like 2*index+1 and 2*index+2 to child nodes?

Assigning indices this way simulates the position of nodes in a full binary tree, allowing us to calculate width by subtracting the first and last indices at each level (see execution_table rows 2-4).

Why do we subtract startIndex from endIndex and add 1 to get width?

Because indices start at 0, width is the count of positions between start and end inclusive, so width = endIndex - startIndex + 1 (see execution_table columns Width Calculated).

What happens if the tree is skewed and some nodes are missing?

The indexing still accounts for missing nodes by skipping indices, so width reflects the maximum gap including nulls (see how indices jump in execution_table rows 3 and 4).

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the width calculated for that level?

A 2

B 3

C 1

D 4

💡 Hint

Check the 'Width Calculated' column at Step 3 in the execution_table

At which step does the maxWidth first update to 3?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Max Width' column in execution_table and find when it changes to 3

If the root had no children, what would be the maxWidth returned?

A 0

B 1

C 2

D Undefined

💡 Hint

Refer to variable_tracker and execution_table Step 1 where only root node is present

Concept Snapshot

Maximum Width of Binary Tree:
- Use BFS level order traversal with a queue
- Assign indices to nodes: left = 2*index+1, right = 2*index+2
- Width at level = last_index - first_index + 1
- Track max width across all levels
- Return max width after traversal

Full Transcript

To find the maximum width of a binary tree, we start at the root and use a queue to traverse level by level. Each node is assigned an index representing its position if the tree were complete. For each level, we calculate the width by subtracting the first node's index from the last node's index and adding one. We update the maximum width found so far if the current level's width is larger. We continue until all levels are processed and then return the maximum width. This method accounts for missing nodes by using indices that reflect their positions in a full binary tree.

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