BST Insert Operation in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after inserting nodes in BST

What is the printed state of the BST after inserting the values 10, 5, 15, 3, 7 in that order? The BST is printed using in-order traversal (left -> root -> right).

DSA Typescript

class Node {
  value: number;
  left: Node | null = null;
  right: Node | null = null;
  constructor(value: number) {
    this.value = value;
  }
}

class BST {
  root: Node | null = null;

  insert(value: number) {
    const newNode = new Node(value);
    if (!this.root) {
      this.root = newNode;
      return;
    }
    let current = this.root;
    while (true) {
      if (value < current.value) {
        if (!current.left) {
          current.left = newNode;
          return;
        }
        current = current.left;
      } else {
        if (!current.right) {
          current.right = newNode;
          return;
        }
        current = current.right;
      }
    }
  }

  inOrder(node: Node | null, result: number[] = []) {
    if (!node) return result;
    this.inOrder(node.left, result);
    result.push(node.value);
    this.inOrder(node.right, result);
    return result;
  }
}

const bst = new BST();
bst.insert(10);
bst.insert(5);
bst.insert(15);
bst.insert(3);
bst.insert(7);
console.log(bst.inOrder(bst.root));

A [10, 5, 15, 3, 7]

B [3, 5, 7, 10, 15]

C [15, 10, 7, 5, 3]

D [3, 7, 5, 10, 15]

💡 Hint

Remember that in-order traversal of a BST prints values in sorted order.

In-order traversal visits left subtree, then root, then right subtree. After inserting 10, 5, 15, 3, 7, the BST in-order traversal outputs sorted values: 3, 5, 7, 10, 15.

❓ Predict Output

intermediate

BST structure after inserting duplicate value

What is the in-order traversal output after inserting values 20, 10, 30, 20 into an empty BST? Assume duplicates are inserted to the right subtree.

DSA Typescript

class Node {
  value: number;
  left: Node | null = null;
  right: Node | null = null;
  constructor(value: number) {
    this.value = value;
  }
}

class BST {
  root: Node | null = null;

  insert(value: number) {
    const newNode = new Node(value);
    if (!this.root) {
      this.root = newNode;
      return;
    }
    let current = this.root;
    while (true) {
      if (value < current.value) {
        if (!current.left) {
          current.left = newNode;
          return;
        }
        current = current.left;
      } else {
        if (!current.right) {
          current.right = newNode;
          return;
        }
        current = current.right;
      }
    }
  }

  inOrder(node: Node | null, result: number[] = []) {
    if (!node) return result;
    this.inOrder(node.left, result);
    result.push(node.value);
    this.inOrder(node.right, result);
    return result;
  }
}

const bst = new BST();
bst.insert(20);
bst.insert(10);
bst.insert(30);
bst.insert(20);
console.log(bst.inOrder(bst.root));

A [10, 20, 20, 30]

B [10, 20, 30]

C [20, 10, 20, 30]

D [10, 30, 20, 20]

💡 Hint

Duplicates go to the right subtree, so the second 20 is placed as right child of the first 20.

Inserting duplicate 20 goes to the right subtree of the first 20. In-order traversal visits left, root, right, so output is [10, 20, 20, 30].

🔧 Debug

advanced

Identify the bug in BST insert method

What error will occur when running this BST insert code if we try to insert 5 into the BST with root 10 and left child 5?

DSA Typescript

class Node {
  value: number;
  left: Node | null = null;
  right: Node | null = null;
  constructor(value: number) {
    this.value = value;
  }
}

class BST {
  root: Node | null = null;

  insert(value: number) {
    const newNode = new Node(value);
    if (!this.root) {
      this.root = newNode;
      return;
    }
    let current = this.root;
    while (true) {
      if (value < current.value) {
        if (!current.left) {
          current.left = newNode;
          return;
        }
        current = current.left;
      } else if (value > current.value) {
        if (!current.right) {
          current.right = newNode;
          return;
        }
        current = current.right;
      }
    }
  }
}

const bst = new BST();
bst.root = new Node(10);
bst.root.left = new Node(5);
bst.insert(5);

A No error, insertion succeeds

B TypeError: Cannot read property 'left' of null

C SyntaxError: Unexpected token 'else if'

D Infinite loop causing program to hang

💡 Hint

Check what happens when value equals current.value in the loop.

The insert method does not handle the case when value equals current.value, so the loop never breaks or inserts, causing infinite loop.

❓ Predict Output

advanced

BST structure after multiple inserts and in-order traversal

After inserting the values 50, 30, 70, 20, 40, 60, 80 into an empty BST, what is the in-order traversal output?

DSA Typescript

class Node {
  value: number;
  left: Node | null = null;
  right: Node | null = null;
  constructor(value: number) {
    this.value = value;
  }
}

class BST {
  root: Node | null = null;

  insert(value: number) {
    const newNode = new Node(value);
    if (!this.root) {
      this.root = newNode;
      return;
    }
    let current = this.root;
    while (true) {
      if (value < current.value) {
        if (!current.left) {
          current.left = newNode;
          return;
        }
        current = current.left;
      } else {
        if (!current.right) {
          current.right = newNode;
          return;
        }
        current = current.right;
      }
    }
  }

  inOrder(node: Node | null, result: number[] = []) {
    if (!node) return result;
    this.inOrder(node.left, result);
    result.push(node.value);
    this.inOrder(node.right, result);
    return result;
  }
}

const bst = new BST();
[50, 30, 70, 20, 40, 60, 80].forEach(v => bst.insert(v));
console.log(bst.inOrder(bst.root));

A [80, 70, 60, 50, 40, 30, 20]

B [50, 30, 20, 40, 70, 60, 80]

C [20, 30, 40, 50, 60, 70, 80]

D [20, 40, 30, 60, 80, 70, 50]

💡 Hint

In-order traversal prints BST values in ascending order.

The in-order traversal visits nodes in ascending order: 20, 30, 40, 50, 60, 70, 80.

🧠 Conceptual

expert

Number of nodes in BST after insertions

If you insert the values [15, 10, 20, 10, 25, 15] into an empty BST where duplicates are inserted to the right subtree, how many nodes will the BST contain?

A 6

B 4

C 3

D 5

💡 Hint

Count each inserted value as a node, including duplicates inserted to the right.

All 6 values are inserted, but duplicates 10 and 15 are inserted as new nodes to the right subtree, so total nodes = 6.

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 keeping the BST property intact

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 property

   A BST keeps nodes ordered so that left child is smaller and right child is larger than the parent.

2. Step 2: Purpose of insert operation

   Insert adds a new node in the correct position to maintain this order.

3. Final Answer:

   To add a new node while keeping the BST property intact -> Option A

4. Quick Check:

   Insert = Add node maintaining order [OK]

Hint: Insert adds nodes without breaking BST order [OK]

Common Mistakes:

• Confusing insert with delete operation
• Thinking insert balances the tree automatically
• Assuming insert finds max or min values

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

easy

A. function insert(root: Node, value: number): void

B. function insert(value: number, root: Node): void

C. function insert(root: Node | null, value: number): Node | null

D. function insert(root: Node, value: string): Node

Solution

1. Step 1: Analyze parameter types

   Insert needs the root node (which can be null if tree is empty) and the value to insert (a number).

2. Step 2: Analyze return type

   It returns the updated root node (Node or null if empty).

3. Final Answer:

   function insert(root: Node | null, value: number): Node | null -> Option C

4. Quick Check:

   Correct params and return type [OK]

Hint: Insert returns updated root, accepts root or null [OK]

Common Mistakes:

• Swapping parameter order
• Using void return type instead of returning Node
• Using wrong value type like string

3. Given the following BST and the insert operation call insert(root, 7), what will be the in-order traversal of the BST after insertion?

Initial BST in-order: 3 -> 5 -> 8 -> 10

Code snippet:

class Node {
  constructor(public val: number, public left: Node | null = null, public right: Node | null = null) {}
}

function insert(root: Node | null, val: number): Node {
  if (!root) return new Node(val);
  if (val < root.val) root.left = insert(root.left, val);
  else root.right = insert(root.right, val);
  return root;
}

medium

A. [3, 5, 8, 7, 10]

B. [3, 5, 8, 10, 7]

C. [7, 3, 5, 8, 10]

D. [3, 5, 7, 8, 10]

Solution

1. Step 1: Locate insertion point for 7

   7 is greater than 5 but less than 8, so it goes to the left of 8.

2. Step 2: Perform in-order traversal after insertion

   In-order visits left, root, right: 3, 5, 7, 8, 10.

3. Final Answer:

   [3, 5, 7, 8, 10] -> Option D

4. Quick Check:

   In-order sorted after insert [OK]

Hint: Insert left if smaller, right if larger [OK]

Common Mistakes:

• Placing 7 incorrectly after 8
• Confusing in-order traversal order
• Ignoring recursive insert logic

4. The following TypeScript code attempts to insert a value into a BST but has a bug. What is the error?

function insert(root: Node | null, val: number): Node {
  if (root === null) {
    return new Node(val);
  }
  if (val <= root.val) {
    insert(root.left, val);
  } else {
    insert(root.right, val);
  }
  return root;
}

medium

A. The base case is missing

B. The recursive calls do not update root.left or root.right

C. The function returns void instead of Node

D. The comparison operator should be < instead of <=

Solution

1. Step 1: Check recursive calls

   The calls to insert do not assign the result back to root.left or root.right, so tree structure is not updated.

2. Step 2: Correct recursive assignment

   Should be root.left = insert(root.left, val) or root.right = insert(root.right, val) to update links.

3. Final Answer:

   The recursive calls do not update root.left or root.right -> Option B

4. Quick Check:

   Must assign recursive insert result to child [OK]

Hint: Always assign recursive insert result to child node [OK]

Common Mistakes:

• Forgetting to assign recursive call result
• Misunderstanding base case presence
• Changing comparison operator unnecessarily

5. You have a BST with nodes 10, 5, 15, 3, 7, 12, 18. You want to insert 6. Which sequence of nodes will you visit during the insert operation?

hard

A. [10, 5, 7]

B. [10, 15, 12]

C. [10, 5, 3]

D. [10, 15, 18]

Solution

1. Step 1: Start at root 10

   6 is less than 10, so move to left child 5.

2. Step 2: Compare with 5 and 7

   6 is greater than 5, move to right child 7. 6 is less than 7, so insert here.

3. Final Answer:

   [10, 5, 7] -> Option A

4. Quick Check:

   Traverse nodes comparing 6 until insertion point [OK]

Hint: Follow left/right based on comparison until null [OK]

Common Mistakes:

• Visiting wrong subtree based on comparison
• Skipping nodes in path
• Confusing insertion path with traversal order