Create a Binary Tree Manually in DSA Typescript - Build from Scratch

Create a Binary Tree Manually

📖 Scenario: You are building a simple binary tree by manually creating nodes and linking them. This is like making a family tree by adding each person and connecting parents to children.

🎯 Goal: Build a binary tree by creating nodes with values and linking left and right children manually. Then print the root node to see the tree structure.

📋 What You'll Learn

Create a class called TreeNode with value, left, and right properties

Create a root node with value 10

Create left and right child nodes with values 5 and 15

Link the child nodes to the root node

Print the root node to show the tree

💡 Why This Matters

🌍 Real World

Binary trees are used in many places like organizing data, searching quickly, and making decisions step-by-step.

💼 Career

Understanding how to build and link nodes in a tree is a key skill for software developers working with data structures and algorithms.

1

Create the TreeNode class

Create a class called TreeNode with a constructor that takes a number value and sets this.value. Initialize this.left and this.right to null.

DSA Typescript

// Your code here

Hint

Think of each node as a box holding a number and two empty spots for left and right children.

2

Create the root and child nodes

Create a variable called root and set it to a new TreeNode with value 10. Then create two more variables called leftChild and rightChild with values 5 and 15 respectively.

DSA Typescript

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

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

// Your code here

Hint

Use the new keyword to create nodes with the values given.

3

Link the child nodes to the root

Set root.left to leftChild and root.right to rightChild to link the children to the root node.

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);
const leftChild = new TreeNode(5);
const rightChild = new TreeNode(15);

// Your code here

Hint

Assign the child nodes to the left and right properties of the root.

4

Print the root node

Use console.log(root) to print the root node and see the tree structure.

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);
const leftChild = new TreeNode(5);
const rightChild = new TreeNode(15);

root.left = leftChild;
root.right = rightChild;

// Your code here

Hint

Use console.log to see the whole tree starting from the root.

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