BST Insert Operation in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A Binary Search Tree is a tree data structure where each node has at most two children. For every node, all nodes in its left subtree have smaller values, and all nodes in its right subtree have larger values.

beginner

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

Compare the new value with the current node's value. If smaller, go to the left child; if larger, go to the right child. Repeat until you find an empty spot to insert the new node.

intermediate

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

Usually, BSTs do not allow duplicate values. The insert operation will ignore the new value or handle duplicates based on specific rules, but commonly it does not insert duplicates.

beginner

Show the step-by-step insertion of 5 into this BST: 10 -> 4 -> 15

Start at root 10: 5 < 10, go left to 4. 5 > 4, go right which is empty. Insert 5 as right child of 4.

intermediate

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

The average time complexity is O(log n) if the tree is balanced, but in the worst case (like a linked list), it can be O(n).

When inserting a new value into a BST, where do you go if the new value is greater than the current node's value?

A To the left child

B To the right child

C Replace the current node

D Stop insertion

In a BST, values greater than the current node go to the right child.

What is the first step when inserting a new node into a BST?

A Compare with the root node

B Insert at the leftmost position

C Insert at the rightmost position

D Delete the root node

Insertion starts by comparing the new value with the root node.

If a BST is unbalanced and looks like a linked list, what is the worst-case time complexity for insertion?

A O(n)

B O(log n)

C O(1)

D O(n log n)

In the worst case, insertion takes O(n) time if the BST is like a linked list.

Which of these is NOT true about BST insertion?

A Insertion compares values to decide direction

B Insertion follows the BST property

C Insertion finds the correct empty spot

D Duplicates are always inserted to the right

Duplicates are usually not inserted or handled specially; they are not always inserted to the right.

After inserting a new node, what property must the BST maintain?

A All left children are greater than the node

B All right children are smaller than the node

C Left subtree nodes are smaller, right subtree nodes are greater

D The tree becomes balanced automatically

BST property requires left subtree nodes to be smaller and right subtree nodes to be greater.

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

Think about walking down the tree comparing values.

Describe the time complexity of inserting a node in a BST and what affects it.

Consider how deep the tree is.

Practice

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

easy

A. To delete a node from the BST

B. To find the height of the BST

C. To add a new node while maintaining BST properties

D. To traverse the BST in order

Solution

1. Step 1: Understand BST insert operation

   The insert operation adds a new node to the BST while keeping the order property intact: left child < parent < right child.

2. Step 2: Differentiate from other operations

   Deleting nodes, finding height, or traversing are different operations and do not describe insertion.

3. Final Answer:

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

4. Quick Check:

   Insert = Add node maintaining order [OK]

Hint: Insert means add node keeping BST order intact [OK]

Common Mistakes:

• Confusing insert with delete operation
• Thinking insert changes tree height only
• Mixing insert with traversal methods

2. Which of the following is the correct syntax to insert a value val into a BST using a recursive function in C++?

easy

A. root = insert(root, val);

B. insert(root, val);

C. root.insert(val);

D. insert(val, root);

Solution

1. Step 1: Understand recursive insert signature

   Recursive insert usually returns the new root after insertion, so assignment like root = insert(root, val); is needed.

2. Step 2: Check other options

   Calling insert(root, val); without assignment ignores updated root. root.insert(val); assumes method on node object which is uncommon. insert(val, root); has wrong parameter order.

3. Final Answer:

   root = insert(root, val); -> Option A

4. Quick Check:

   Assign returned root after insert [OK]

Hint: Assign root = insert(root, val) for recursive insertion [OK]

Common Mistakes:

• Not assigning the returned root after insertion
• Using wrong parameter order in insert call
• Assuming insert is a method of node object

3. Consider the following C++ code inserting values into an empty BST:

struct Node {
  int data;
  Node* left;
  Node* right;
  Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* insert(Node* root, int val) {
  if (!root) return new Node(val);
  if (val < root->data) root->left = insert(root->left, val);
  else root->right = insert(root->right, val);
  return root;
}

int main() {
  Node* root = nullptr;
  root = insert(root, 10);
  root = insert(root, 5);
  root = insert(root, 15);
  root = insert(root, 12);
  return 0;
}

What is the structure of the BST after these insertions?

medium

A. 10 -> 5 -> 12 (left), 10 -> 15 -> null (right subtree)

B. 10 -> 15 -> null (left), 10 -> 5 -> 12 (right subtree)

C. 10 -> 12 -> 5 (left), 10 -> 15 -> null (right subtree)

D. 10 -> 5 -> null (left), 10 -> 15 -> 12 (right subtree)

Solution

1. Step 1: Insert 10, 5, 15, 12 step-by-step

   Start with empty root. Insert 10 as root. Insert 5 goes left of 10. Insert 15 goes right of 10. Insert 12 goes left of 15.

2. Step 2: Visualize BST structure

   Root 10 has left child 5 (no children), right child 15 with left child 12 (no right child).

3. Final Answer:

   10 -> 5 -> null (left), 10 -> 15 -> 12 (right subtree) -> Option D

4. Quick Check:

   Insert order builds left 5, right 15 with left 12 [OK]

Hint: Insert smaller left, larger right; 12 is left child of 15 [OK]

Common Mistakes:

• Placing 12 as right child of 15 instead of left
• Swapping left and right children of 10
• Ignoring null children in output

4. The following code attempts to insert a value into a BST but causes a runtime error:

Node* insert(Node* root, int val) {
  if (root == nullptr) {
    Node* newNode = new Node(val);
  }
  if (val < root->data) {
    root->left = insert(root->left, val);
  } else {
    root->right = insert(root->right, val);
  }
  return root;
}

What is the error and how to fix it?

medium

A. Missing return after creating new node; add return newNode;

B. Incorrect comparison operator; use <= instead of <

C. Should not assign to root->left or root->right

D. Node constructor is missing initialization

Solution

1. Step 1: Identify missing return in base case

   When root is nullptr, new node is created but not returned, so function continues and dereferences nullptr causing runtime error.

2. Step 2: Fix by returning new node immediately

   Add return newNode; inside the if block to return the newly created node and stop further execution.

3. Final Answer:

   Missing return after creating new node; add return newNode; -> Option A

4. Quick Check:

   Return new node on nullptr base case [OK]

Hint: Always return new node immediately when root is nullptr [OK]

Common Mistakes:

• Not returning new node causes nullptr dereference
• Changing comparison operator unnecessarily
• Misunderstanding pointer assignments

5. You have a BST with nodes 20, 10, 30, 25, 40. You want to insert 15 and 35. After insertion, which of the following is the correct in-order traversal of the BST?

hard

A. [15, 10, 20, 25, 30, 35, 40]

B. [10, 15, 20, 25, 30, 35, 40]

C. [10, 20, 15, 25, 30, 35, 40]

D. [10, 15, 20, 30, 25, 35, 40]

Solution

1. Step 1: Insert 15 and 35 into BST

   Insert 15: less than 20, greater than 10, goes right of 10. Insert 35: less than 40, greater than 30, goes right of 30.

2. Step 2: Perform in-order traversal

   In-order traversal visits nodes in ascending order: left subtree, root, right subtree. The order is 10, 15, 20, 25, 30, 35, 40.

3. Final Answer:

   [10, 15, 20, 25, 30, 35, 40] -> Option B

4. Quick Check:

   In-order traversal = sorted nodes [OK]

Hint: In-order traversal of BST is sorted list [OK]

Common Mistakes:

• Placing 15 or 35 in wrong subtree
• Confusing traversal order
• Listing nodes unsorted in output