BST Insert Operation in DSA Java - Build from Scratch

BST Insert Operation

📖 Scenario: You are building a simple phone book application that stores phone numbers in a Binary Search Tree (BST) to keep them sorted for quick search and insertion.

🎯 Goal: Learn how to insert new phone numbers into a BST while maintaining its sorted property.

📋 What You'll Learn

Create a BST node class with integer data and left/right child references

Create a BST class with an insert method to add new nodes

Insert multiple phone numbers into the BST

Print the BST in-order to show sorted phone numbers

💡 Why This Matters

🌍 Real World

Phone books, contact lists, and databases use BSTs to keep data sorted for fast insertion and search.

💼 Career

Understanding BST insertions is key for software engineers working on efficient data storage and retrieval systems.

1

Create BST Node Class

Create a class called Node with an integer variable data, and two Node references called left and right. Add a constructor that sets data and initializes left and right to null.

DSA Java

// Your code here

Hint

Think of Node as a box holding a number and links to two other boxes.

2

Create BST Class with Root

Create a class called BST with a Node variable called root. Initialize root to null in the constructor.

DSA Java

class Node {
    int data;
    Node left, right;

    Node(int data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

// Your code here

Hint

The BST class holds the starting point of your tree called root.

3

Add Insert Method to BST

Inside the BST class, add a method called insert that takes an integer data and inserts it into the BST. Use a helper method insertRec that takes a Node and data, and returns the updated Node. Insert smaller values to the left and larger or equal values to the right.

DSA Java

class Node {
    int data;
    Node left, right;

    Node(int data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

class BST {
    Node root;

    BST() {
        this.root = null;
    }

    // Add insert and insertRec methods below
    // Your code here
}

Hint

Use recursion to find the right place to add the new number.

4

Insert Numbers and Print In-Order

In the BST class, add a method called inOrder that prints the BST nodes in ascending order separated by spaces. In the main method, create a BST object, insert these numbers in order: 50, 30, 70, 20, 40, 60, 80. Then call inOrder to print the sorted numbers.

DSA Java

class Node {
    int data;
    Node left, right;

    Node(int data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

class BST {
    Node root;

    BST() {
        this.root = null;
    }

    void insert(int data) {
        root = insertRec(root, data);
    }

    Node insertRec(Node root, int data) {
        if (root == null) {
            root = new Node(data);
            return root;
        }
        if (data < root.data) {
            root.left = insertRec(root.left, data);
        } else {
            root.right = insertRec(root.right, data);
        }
        return root;
    }

    // Add inOrder method and main method below
    // Your code here
}

Hint

In-order traversal prints the numbers in sorted order.

Practice

1. What is the main purpose of the insert operation in a Binary Search Tree (BST)?

easy

A. To add a new node while maintaining BST properties

B. To delete a node from the BST

C. To find the maximum value in the BST

D. To balance the BST after insertion

Solution

1. Step 1: Understand BST insert operation

   The insert operation adds a new node to the BST without breaking the BST property where left child nodes are smaller and right child nodes are larger.

2. Step 2: Identify the purpose

   Insertion is specifically for adding new nodes, not deleting or balancing the tree.

3. Final Answer:

   To add a new node while maintaining BST properties -> Option A

4. Quick Check:

   Insert = Add node [OK]

Hint: Insert means add node keeping order [OK]

Common Mistakes:

• Confusing insert with delete operation
• Thinking insert balances the tree automatically
• Assuming insert finds max value

2. Which of the following is the correct Java method signature for inserting a value into a BST node?

easy

A. public void insert(int value)

B. public boolean insert(int value, Node root)

C. public int insert(Node root, int value)

D. public Node insert(Node root, int value)

Solution

1. Step 1: Analyze method parameters and return type

   Insertion usually requires the root node and the value to insert. It returns the updated root node after insertion.

2. Step 2: Match signature with common BST insert pattern

   The method returns a Node and takes Node root and int value as parameters, matching public Node insert(Node root, int value).

3. Final Answer:

   public Node insert(Node root, int value) -> Option D

4. Quick Check:

   Insert method returns Node with root and value [OK]

Hint: Insert method returns updated root node [OK]

Common Mistakes:

• Using void return type instead of Node
• Swapping parameter order incorrectly
• Returning int instead of Node

3. What will be the inorder traversal output after inserting 15 into the BST with nodes 10, 5, 20?

medium

A. [5, 15, 10, 20]

B. [10, 5, 15, 20]

C. [5, 10, 15, 20]

D. [20, 15, 10, 5]

Solution

1. Step 1: Insert 15 into BST

   Start at root 10: 15 > 10, go right to 20. 15 < 20, insert as left child of 20.

2. Step 2: Perform inorder traversal

   Inorder visits left, root, right: left subtree [5], root 10, left child of 20 is 15, then 20. Result: [5, 10, 15, 20]

3. Final Answer:

   [5, 10, 15, 20] -> Option C

4. Quick Check:

   Inorder after insert = sorted list [OK]

Hint: Inorder traversal of BST is sorted [OK]

Common Mistakes:

• Inserting 15 as right child of 10 incorrectly
• Confusing preorder with inorder traversal
• Ignoring BST property during insertion

4. Identify the error in this BST insert method snippet:

public Node insert(Node root, int val) {
  if (root == null) {
    root = new Node(val);
  } else if (val < root.val) {
    insert(root.left, val);
  } else {
    insert(root.right, val);
  }
  return root;
}

medium

A. Incorrect base case for null root

B. Missing return when recursively inserting left or right

C. Using assignment instead of comparison in if condition

D. Returning void instead of Node

Solution

1. Step 1: Check recursive calls

   Recursive calls to insert(root.left, val) and insert(root.right, val) do not assign the returned Node back to root.left or root.right.

2. Step 2: Identify fix

   Must assign: root.left = insert(root.left, val); and root.right = insert(root.right, val); to update tree links.

3. Final Answer:

   Missing return when recursively inserting left or right -> Option B

4. Quick Check:

   Assign recursive insert result to child [OK]

Hint: Assign recursive insert result to child node [OK]

Common Mistakes:

• Not assigning recursive insert result to left/right child
• Confusing base case logic
• Returning wrong type

5. Given a BST with nodes inserted in this order: 30, 20, 40, 10, 25, 35, 50, what will be the structure after inserting 22? Choose the correct inorder traversal output.

hard

A. [10, 20, 22, 25, 30, 35, 40, 50]

B. [10, 20, 25, 22, 30, 35, 40, 50]

C. [10, 22, 20, 25, 30, 35, 40, 50]

D. [10, 20, 25, 30, 22, 35, 40, 50]

Solution

1. Step 1: Insert 22 into BST

   Start at root 30: 22 < 30 go left to 20; 22 > 20 go right to 25; 22 < 25 insert as left child of 25.

2. Step 2: Perform inorder traversal

   Inorder visits nodes in sorted order: 10, 20, 22, 25, 30, 35, 40, 50.

3. Final Answer:

   [10, 20, 22, 25, 30, 35, 40, 50] -> Option A

4. Quick Check:

   Inorder traversal is sorted list including 22 [OK]

Hint: Insert by BST rules, inorder is sorted [OK]

Common Mistakes:

• Inserting 22 in wrong subtree
• Misordering nodes in traversal
• Confusing preorder with inorder