Create a Binary Tree Manually in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of the manual binary tree traversal?

Consider the following TypeScript code that creates a simple binary tree manually and performs an inorder traversal (left, root, right). What will be the printed output?

DSA Typescript

class TreeNode {
  value: number;
  left: TreeNode | null;
  right: TreeNode | null;

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

const root = new TreeNode(10);
root.left = new TreeNode(5);
root.right = new TreeNode(15);
root.left.left = new TreeNode(3);
root.left.right = new TreeNode(7);

function inorder(node: TreeNode | null): void {
  if (!node) return;
  inorder(node.left);
  console.log(node.value);
  inorder(node.right);
}

inorder(root);

A 10\n5\n15\n3\n7

B 5\n3\n7\n10\n15

C 3\n5\n7\n10\n15

D 3\n7\n5\n10\n15

💡 Hint

Remember inorder traversal visits left child, then root, then right child.

Inorder traversal visits nodes in the order: left subtree, root, right subtree. So the output is 3, 5, 7, 10, 15.

🧠 Conceptual

intermediate

How many nodes are in this manually created binary tree?

Given the following manual binary tree creation code, how many nodes does the tree contain?

DSA Typescript

class TreeNode {
  value: number;
  left: TreeNode | null;
  right: TreeNode | null;

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

const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.right.right = new TreeNode(6);

A 4

B 6

C 7

D 5

💡 Hint

Count each new TreeNode created and assigned.

Nodes created: 1 (root), 2, 3, 4, 5, 6. Total 6 nodes.

🔧 Debug

advanced

What error occurs when running this manual binary tree code?

Examine the following TypeScript code that attempts to create a binary tree. What error will it produce when run?

DSA Typescript

class TreeNode {
  value: number;
  left: TreeNode | null;
  right: TreeNode | null;

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

const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.right.left.left = new TreeNode(6);

A TypeError: Cannot set property 'left' of null

B SyntaxError: Unexpected token '.'

C No error, tree created successfully

D ReferenceError: root.right.left is not defined

💡 Hint

Check if all intermediate nodes exist before assigning children.

root.right.left is null by default, so trying to access .left on it causes a TypeError.

📝 Syntax

advanced

Which option correctly creates a manual binary tree node with left and right children?

Choose the correct TypeScript code snippet that creates a binary tree node with value 10 and assigns left child 5 and right child 15.

A

const node = new TreeNode(10);
node.left = new TreeNode(5);
node.right = new TreeNode(15);

B

const node = TreeNode(10);
node.left = TreeNode(5);
node.right = TreeNode(15);

C

const node = new TreeNode(10)
node.left = new TreeNode(5)
node.right = new TreeNode(15)

D

const node = new TreeNode(10);
node.left = 5;
node.right = 15;

💡 Hint

Remember to use new keyword and assign TreeNode objects to left and right.

Option A correctly uses new and assigns TreeNode instances to left and right.

🚀 Application

expert

What is the preorder traversal output of this manually created binary tree?

Given the following manual binary tree creation, what is the output of a preorder traversal (root, left, right)?

DSA Typescript

class TreeNode {
  value: number;
  left: TreeNode | null;
  right: TreeNode | null;

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

const root = new TreeNode(8);
root.left = new TreeNode(3);
root.right = new TreeNode(10);
root.left.left = new TreeNode(1);
root.left.right = new TreeNode(6);
root.left.right.left = new TreeNode(4);
root.left.right.right = new TreeNode(7);
root.right.right = new TreeNode(14);
root.right.right.left = new TreeNode(13);

function preorder(node: TreeNode | null): void {
  if (!node) return;
  console.log(node.value);
  preorder(node.left);
  preorder(node.right);
}

preorder(root);

A 1\n3\n4\n6\n7\n8\n10\n13\n14

B 3\n1\n6\n4\n7\n8\n10\n14\n13

C 8\n10\n14\n13\n3\n6\n4\n7\n1

D 8\n3\n1\n6\n4\n7\n10\n14\n13

💡 Hint

Preorder visits root first, then left subtree, then right subtree.

Preorder traversal visits nodes in order: root, left, right. So output is 8, 3, 1, 6, 4, 7, 10, 14, 13.

Practice

1. What is the main purpose of manually creating a binary tree in TypeScript?

easy

A. To avoid using classes or objects

B. To automatically sort data without coding

C. To understand how nodes connect and form the tree structure

D. To create a linked list instead of a tree

Solution

1. Step 1: Understand binary tree structure

   A binary tree consists of nodes where each node has at most two children, left and right.

2. Step 2: Purpose of manual creation

   Manually creating a binary tree helps visualize and understand how nodes connect to form the tree.

3. Final Answer:

   To understand how nodes connect and form the tree structure -> Option C

4. Quick Check:

   Manual creation = Understand tree structure [OK]

Hint: Manual creation helps visualize node connections [OK]

Common Mistakes:

• Thinking it automatically sorts data
• Confusing binary tree with linked list
• Assuming no objects or classes are needed

2. Which of the following is the correct way to define a binary tree node class in TypeScript?

easy

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

B. class Node { value: number; left: number; right: number; constructor(value: number) { this.value = value; this.left = 0; this.right = 0; } }

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

D. class Node { value: number; left: any; right: any; constructor() { this.value = 0; } }

Solution

1. Step 1: Check node properties and types

   A binary tree node should have a value and left/right children which can be nodes or null.

2. Step 2: Verify constructor initialization

   Constructor must initialize value and set left and right to null initially.

3. Final Answer:

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

4. Quick Check:

   Node class with nullable children = class Node { value: number; left: Node | null; right: Node | null; constructor(value: number) { this.value = value; this.left = null; this.right = null; } } [OK]

Hint: Children must be Node or null, initialized in constructor [OK]

Common Mistakes:

• Using number type for left/right instead of Node or null
• Not initializing left and right to null
• Omitting constructor parameters

3. Given the following TypeScript code to create a binary tree manually, what will be the output of console.log(root.left?.value)?

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

const root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);

medium

A. 5

B. 10

C. 15

D. undefined

Solution

1. Step 1: Understand tree structure from code

   Root node has value 10, left child is node with value 5, right child is node with value 15.

2. Step 2: Access left child's value

   root.left?.value accesses left node's value safely, which is 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   root.left.value = 5 [OK]

Hint: Access left child value directly with root.left?.value [OK]

Common Mistakes:

• Confusing root value with left child value
• Expecting right child value instead
• Not using optional chaining and getting error

4. Identify the error in the following TypeScript code that tries to create a binary tree node manually:

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

const root = new Node(10);
root.left = 5;
root.right = new Node(15);

medium

A. Missing constructor parameters in Node class

B. Using wrong class name for Node

C. Not initializing root.right

D. Assigning number 5 to root.left instead of a Node or null

Solution

1. Step 1: Check type of root.left assignment

   root.left is declared as Node or null, but assigned number 5 which is invalid.

2. Step 2: Confirm correct assignment type

   root.left should be assigned a Node instance or null, not a number.

3. Final Answer:

   Assigning number 5 to root.left instead of a Node or null -> Option D

4. Quick Check:

   Type mismatch in root.left assignment = Assigning number 5 to root.left instead of a Node or null [OK]

Hint: Children must be Node or null, not a number [OK]

Common Mistakes:

• Assigning primitive values to node children
• Ignoring type errors in assignments
• Assuming number can be assigned directly

5. You want to manually create a binary tree with root value 20, left child 10, and right child 30. Which code snippet correctly creates this tree and prints the root, left, and right values in order?

hard

A. const root = new Node(20);
root.left = 10;
root.right = 30;
console.log(root.value, root.left, root.right);

B. const root = new Node(20);
root.left = new Node(10);
root.right = new Node(30);
console.log(root.value, root.left?.value, root.right?.value);

C. const root = new Node(20);
root.left = new Node(30);
root.right = new Node(10);
console.log(root.value, root.left?.value, root.right?.value);

D. const root = new Node(20);
root.left = null;
root.right = null;
console.log(root.left?.value, root.value, root.right?.value);

Solution

1. Step 1: Create root and children nodes correctly

   Root node with value 20, left child 10, right child 30 must be Node instances.

2. Step 2: Print values in order root, left, right

   console.log prints root.value, root.left?.value, root.right?.value which outputs 20 10 30.

3. Final Answer:

   const root = new Node(20); root.left = new Node(10); root.right = new Node(30); console.log(root.value, root.left?.value, root.right?.value); -> Option B

4. Quick Check:

   Correct node creation and print order = const root = new Node(20); root.left = new Node(10); root.right = new Node(30); console.log(root.value, root.left?.value, root.right?.value); [OK]

Hint: Create nodes with new, assign children as nodes, print values [OK]

Common Mistakes:

• Assigning numbers directly to children
• Swapping left and right child values
• Printing children before root or using null children