Maximum Width of Binary Tree in DSA Go - Time & Space Complexity

Time Complexity: Maximum Width of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the maximum width of a binary tree grows as the tree gets bigger.

Specifically, how does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func widthOfBinaryTree(root *TreeNode) int {
    if root == nil {
        return 0
    }
    maxWidth := 0
    queue := []struct{
        node *TreeNode
        index int
    }{{root, 0}}

    for len(queue) > 0 {
        size := len(queue)
        start := queue[0].index
        end := queue[size-1].index
        if end - start + 1 > maxWidth {
            maxWidth = end - start + 1
        }
        for i := 0; i < size; 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 level-order traversal, tracking node positions.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop runs once per level of the tree, and the inner loop processes every node at that level.
• How many times: Each node is visited exactly once during the traversal.

How Execution Grows With Input

As the number of nodes (n) increases, the code visits each node once, so the total work grows directly with n.

Input Size (n)	Approx. Operations
10	About 10 visits and checks
100	About 100 visits and checks
1000	About 1000 visits and checks

Pattern observation: The work grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum width grows in direct proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "The time complexity depends on the maximum width of the tree, not the total nodes."

[OK] Correct: Even if the width is large, the algorithm must visit every node once, so total nodes determine the time, not just the width.

Interview Connect

Understanding how tree traversal time grows helps you explain your approach clearly and shows you know how to analyze algorithms well.

Self-Check

"What if we used recursion instead of a queue for level-order traversal? How would the time complexity change?"

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