BST Insert Operation in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the given value.

DSA Java

Node newNode = new Node([1]);

Drag options to blanks, or click blank then click option'

Anode

Bvalue

Croot

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' or 'root' instead of the value.

Using 'null' which does not create a node with the value.

We create a new node using the value to insert into the BST.

2fill in blank

medium

Complete the code to check if the root is empty before inserting.

DSA Java

if (root == [1]) {
    root = newNode;
    return root;
}

Drag options to blanks, or click blank then click option'

AnewNode

Bvalue

Cnull

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if root equals newNode instead of null.

Using root itself in the condition.

We check if root is null to know if the tree is empty.

3fill in blank

hard

Fix the error in the recursive insert call to the left subtree.

DSA Java

if (value < root.data) {
    root.left = [1](root.left, value);
}

Drag options to blanks, or click blank then click option'

AaddNode

BinsertNode

CinsertBST

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong method names like insertNode or addNode.

Forgetting to assign the result back to root.left.

The method name is 'insert' for recursive calls.

4fill in blank

hard

Fill both blanks to complete the recursive insert call to the right subtree.

DSA Java

if (value > root.data) {
    root.[1] = insert(root.[2], value);
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Croot

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' for the right subtree.

Using 'root' or 'node' which are incorrect here.

We insert into the right subtree and assign back to root.right.

5fill in blank

hard

Fill all three blanks to complete the insert method signature and return statement.

DSA Java

public static Node [1](Node [2], int [3]) {
    // method body
    return root;
}

Drag options to blanks, or click blank then click option'

Ainsert

Broot

Cvalue

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong method names like 'add'.

Using incorrect parameter names.

The method is named 'insert', takes 'root' node and 'value' to insert.

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