BST Insert Operation in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A BST is a tree where each node has at most two children. The left child's value is less than the parent node's value, and the right child's value is greater.

beginner

What is the main rule to follow when inserting a new value into a BST?

Compare the new value with the current node. If smaller, go left; if larger, go right. Insert the new node where a null child is found.

intermediate

Why do we insert nodes recursively in a BST?

Recursion helps to move down the tree easily until the correct empty spot is found for the new node.

intermediate

What happens if you try to insert a value already present in the BST?

Usually, duplicates are not inserted. The insert operation can ignore or handle duplicates based on the implementation.

beginner

Show the order of nodes after inserting 5, 3, 7 into an empty BST.

The BST will look like: 5 -> left child 3, right child 7. Printed as: 3 -> 5 -> 7 (in-order traversal).

When inserting a value into a BST, if the value is less than the current node, where do you go?

A Left subtree

B Right subtree

C Replace current node

D Stop insertion

In BST, smaller values go to the left subtree.

What is the base case for the recursive BST insert function?

A When the current node has two children

B When the current node is null

C When the value equals the current node

D When the tree is empty

Insertion happens when we find a null spot to place the new node.

If you insert values 10, 5, 15 in order into an empty BST, what is the root node?

A 5

B 15

C None

D 10

The first inserted value becomes the root.

What is the time complexity of inserting a node in a balanced BST?

A O(log n)

B O(n)

C O(1)

D O(n log n)

Balanced BST insertion takes O(log n) time.

Which traversal order prints BST nodes in sorted order?

A Pre-order

B Post-order

C In-order

D Level-order

In-order traversal visits nodes in ascending order.

Explain how to insert a new value into a Binary Search Tree.

Think about how you decide where to put the new value by comparing it with existing nodes.

Describe what happens step-by-step when inserting 8 into this BST: 10 -> left 5, right 15.

Follow the path from root to the correct empty spot.

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