BST Find Maximum Element in DSA Go - Build from Scratch

BST Find Maximum Element

📖 Scenario: You are working with a simple phone book application that stores contacts in a Binary Search Tree (BST). Each contact has a unique phone number. You want to find the contact with the highest phone number.

🎯 Goal: Build a Go program that creates a BST with given phone numbers, then finds and prints the maximum phone number stored in the BST.

📋 What You'll Learn

Create a BST node structure with an integer phone number

Insert given phone numbers into the BST

Write a function to find the maximum phone number in the BST

Print the maximum phone number found

💡 Why This Matters

🌍 Real World

Phone books, contact lists, and databases often use BSTs to store and quickly find data like phone numbers.

💼 Career

Understanding BSTs and how to find maximum or minimum values is important for software engineers working on efficient data storage and retrieval.

1

Create BST Node Structure and Insert Function

Create a struct called Node with an integer field phoneNumber and two pointers left and right to Node. Then write a function insert(root *Node, phoneNumber int) *Node that inserts a phone number into the BST and returns the root node.

DSA Go

package main

// Your code here

Hint

Think of Node as a contact with a phone number and links to smaller and larger numbers.

2

Insert Given Phone Numbers into BST

Create a variable root of type *Node and insert these phone numbers into the BST in this order: 5551234, 5555678, 5550000, 5559999, 5554321.

DSA Go

package main

type Node struct {
    phoneNumber int
    left, right *Node
}

func insert(root *Node, phoneNumber int) *Node {
    if root == nil {
        return &Node{phoneNumber: phoneNumber}
    }
    if phoneNumber < root.phoneNumber {
        root.left = insert(root.left, phoneNumber)
    } else {
        root.right = insert(root.right, phoneNumber)
    }
    return root
}

func main() {
    var root *Node
    // Insert phone numbers into BST below
    // Your code here
}

Hint

Use the insert function to add each phone number to the BST.

3

Write Function to Find Maximum Phone Number

Write a function findMax(root *Node) int that returns the maximum phone number in the BST by following the right child nodes until none remain.

DSA Go

package main

type Node struct {
    phoneNumber int
    left, right *Node
}

func insert(root *Node, phoneNumber int) *Node {
    if root == nil {
        return &Node{phoneNumber: phoneNumber}
    }
    if phoneNumber < root.phoneNumber {
        root.left = insert(root.left, phoneNumber)
    } else {
        root.right = insert(root.right, phoneNumber)
    }
    return root
}

func main() {
    var root *Node
    root = insert(root, 5551234)
    root = insert(root, 5555678)
    root = insert(root, 5550000)
    root = insert(root, 5559999)
    root = insert(root, 5554321)
}

// Write findMax function below
// Your code here

Hint

Keep moving to the right child until there is no right child left, then return that node's phone number.

4

Print the Maximum Phone Number

In the main function, call findMax(root) and print the returned maximum phone number using fmt.Println. Import the fmt package at the top.

DSA Go

package main

import "fmt"

type Node struct {
    phoneNumber int
    left, right *Node
}

func insert(root *Node, phoneNumber int) *Node {
    if root == nil {
        return &Node{phoneNumber: phoneNumber}
    }
    if phoneNumber < root.phoneNumber {
        root.left = insert(root.left, phoneNumber)
    } else {
        root.right = insert(root.right, phoneNumber)
    }
    return root
}

func findMax(root *Node) int {
    current := root
    for current.right != nil {
        current = current.right
    }
    return current.phoneNumber
}

func main() {
    var root *Node
    root = insert(root, 5551234)
    root = insert(root, 5555678)
    root = insert(root, 5550000)
    root = insert(root, 5559999)
    root = insert(root, 5554321)
    // Print the maximum phone number below
    // Your code here
}

Hint

Use fmt.Println(findMax(root)) or assign to a variable then print it.

Practice

1. In a Binary Search Tree (BST), where is the maximum element always located?

easy

A. At any leaf node

B. At the rightmost node

C. At the root node

D. At the leftmost node

Solution

1. Step 1: Understand BST property

   In a BST, all nodes to the right of a node have greater values.

2. Step 2: Locate maximum element

   The maximum element is found by moving to the rightmost node, as it has no greater right child.

3. Final Answer:

   At the rightmost node -> Option B

4. Quick Check:

   Maximum = rightmost node [OK]

Hint: Maximum is always the rightmost node in BST [OK]

Common Mistakes:

• Confusing maximum with root node
• Thinking maximum is at leftmost node
• Assuming maximum can be any leaf

2. Which of the following Go code snippets correctly finds the maximum element in a BST node structure?

easy

A. for current.Left != nil { current = current.Left } return current.Value

B. return root.Value

C. for current.Right != nil { current = current.Right } return current.Value

D. for current != nil { current = current.Left } return current.Value

Solution

1. Step 1: Check loop condition for maximum

   Maximum is found by moving right until no right child exists.

2. Step 2: Verify code correctness

   for current.Right != nil { current = current.Right } return current.Value correctly loops while current.Right is not nil, then returns current.Value.

3. Final Answer:

   for current.Right != nil { current = current.Right } return current.Value -> Option C

4. Quick Check:

   Loop right child until nil = max [OK]

Hint: Loop right child until nil to find max [OK]

Common Mistakes:

• Looping left child instead of right
• Returning root value without traversal
• Loop condition incorrect or infinite

3. What will be the output of the following Go code snippet that finds the maximum element in a BST?

type Node struct {
    Value int
    Left  *Node
    Right *Node
}

func findMax(root *Node) int {
    current := root
    for current.Right != nil {
        current = current.Right
    }
    return current.Value
}

root := &Node{Value: 10}
root.Left = &Node{Value: 5}
root.Right = &Node{Value: 15}
root.Right.Right = &Node{Value: 20}

fmt.Println(findMax(root))

medium

A. 20

B. 15

C. 10

D. 5

Solution

1. Step 1: Trace the BST structure

   Root is 10, right child 15, right-right child 20 (rightmost node).

2. Step 2: Follow findMax function

   Function moves right until current.Right is nil, stopping at node with value 20.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Rightmost node value = 20 [OK]

Hint: Rightmost node value is maximum [OK]

Common Mistakes:

• Returning root value instead of rightmost
• Stopping at first right child
• Confusing left and right pointers

4. The following Go function is intended to find the maximum element in a BST but contains a bug. What is the bug?

func findMax(root *Node) int {
    current := root
    for current.Left != nil {
        current = current.Left
    }
    return current.Value
}

medium

A. All of the above

B. It does not handle empty tree (nil root)

C. It returns the minimum element instead of maximum

D. It incorrectly traverses left children instead of right

Solution

1. Step 1: Analyze traversal direction

   The code loops while current.Left != nil, moving left, which finds minimum, not maximum.

2. Step 2: Check for other issues

   Function does not check if root is nil, which can cause runtime error.

3. Final Answer:

   It incorrectly traverses left children instead of right; does not handle nil root; returns minimum instead of maximum -> Option A

4. Quick Check:

   Left traversal + no nil check = bugs [OK]

Hint: Traverse right, check nil root to avoid bugs [OK]

Common Mistakes:

• Using left instead of right pointer
• Ignoring nil root check
• Assuming code finds maximum correctly

5. Given a BST where some nodes have duplicate values, how would you modify the Go function to find the maximum element considering duplicates are allowed on the right subtree? Choose the best approach.

hard

A. Traverse left while current.Left != nil and current.Left.Value > current.Value

B. Return root.Value directly since duplicates are on right

C. Traverse right while current.Right != nil and current.Right.Value > current.Value

D. Traverse right while current.Right != nil only

Solution

1. Step 1: Understand duplicates in BST

   In a BST allowing duplicates on right, right subtree values are >= current value.

2. Step 2: Best traversal for maximum

   Traverse right while current.Right != nil only. Value checks like in C (>) stop early on equal duplicates before larger values (e.g., 10->10->12).

3. Final Answer:

   Traverse right while current.Right != nil only -> Option D

4. Quick Check:

   Structural rightmost traversal = max even with duplicates [OK]

Hint: Traverse right until nil regardless of duplicates [OK]

Common Mistakes:

• Using strict > comparison, stopping early on equals
• Traversing left instead of right
• Returning root value without traversal