Binary Tree Node Structure in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Binary Tree Node Creation

What will be the printed output of the binary tree node after running the code below?

DSA Typescript

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

const root = new TreeNode(10);
root.left = new TreeNode(5);
root.right = new TreeNode(15);
console.log(root.val + ' -> ' + root.left?.val + ' -> ' + root.right?.val + ' -> null');

A 10 -> 5 -> 15 -> null

B 10 -> null -> null -> null

C 15 -> 5 -> 10 -> null

D 5 -> 10 -> 15 -> null

💡 Hint

Think about how the root node and its children are assigned values.

The root node has value 10, its left child is 5, and right child is 15. The print statement shows these values in order followed by null.

❓ Predict Output

intermediate

Value of Left Child After Node Modification

After running the code below, what is the value of root.left.val?

DSA Typescript

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

const root = new TreeNode(20);
root.left = new TreeNode(10);
root.left.val = 15;

A 15

B 10

C 20

D null

💡 Hint

Look at the line where root.left.val is changed.

The left child's value was initially 10 but then changed to 15 explicitly.

❓ Predict Output

advanced

Output of Inorder Traversal of Binary Tree

What is the output of the inorder traversal of the binary tree created by the code below?

DSA Typescript

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

function inorder(root: TreeNode | null, result: number[] = []): number[] {
  if (root === null) return result;
  inorder(root.left, result);
  result.push(root.val);
  inorder(root.right, result);
  return result;
}

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.right.right = new TreeNode(14);
const output = inorder(root);
console.log(output.join(' -> ') + ' -> null');

A 8 -> 3 -> 1 -> 6 -> 10 -> 14 -> null

B 1 -> 3 -> 6 -> 8 -> 10 -> 14 -> null

C 1 -> 6 -> 3 -> 8 -> 14 -> 10 -> null

D 3 -> 1 -> 6 -> 8 -> 14 -> 10 -> null

💡 Hint

Inorder traversal visits left child, then node, then right child.

Inorder traversal visits nodes in ascending order for binary search trees. The sequence is left subtree (1,3,6), root (8), right subtree (10,14).

🔧 Debug

advanced

Identify the Error in Binary Tree Node Assignment

What error will the following code produce when run?

DSA Typescript

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

const root = new TreeNode(5);
root.left.val = 3;

A No error, runs successfully

B SyntaxError: Unexpected token '.'

C ReferenceError: root.left is not defined

D TypeError: Cannot read property 'val' of null

💡 Hint

Check if root.left is assigned before accessing its val property.

root.left is null by default. Trying to access val on null causes a TypeError.

🧠 Conceptual

expert

Number of Nodes in a Complete Binary Tree of Height h

What is the total number of nodes in a complete binary tree of height h (where height is the number of edges from root to deepest leaf)?

A h + 1

B 2^h - 1

C 2^(h+1) - 1

D h * 2

💡 Hint

Think about how nodes double at each level in a complete binary tree.

A complete binary tree of height h has nodes equal to sum of nodes at each level: 1 + 2 + 4 + ... + 2^h = 2^(h+1) - 1.

Practice

1. What does a typical binary tree node contain in its structure?

easy

A. Only a value and a pointer to the next node

B. A value, a left child node, and a right child node

C. A value and multiple child nodes in a list

D. Only a value without any child nodes

Solution

1. Step 1: Understand binary tree node components

   A binary tree node typically stores a value and references to two children: left and right.

2. Step 2: Compare options with definition

   A value, a left child node, and a right child node correctly lists value, left child, and right child. Others describe linked list or general tree nodes.

3. Final Answer:

   A value, a left child node, and a right child node -> Option B

4. Quick Check:

   Binary tree node = value + left + right [OK]

Hint: Binary tree nodes always have two child pointers [OK]

Common Mistakes:

• Confusing binary tree node with linked list node
• Thinking nodes have multiple children like general trees
• Ignoring child pointers in node structure

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

easy

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

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

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

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

Solution

1. Step 1: Check property types and initialization

   class Node { value: number; left: Node | null; right: Node | null; constructor(value: number) { this.value = value; this.left = null; this.right = null; } } correctly declares left and right as Node or null and initializes them to null in constructor.

2. Step 2: Verify constructor parameters and assignments

   class Node { value: number; left: Node | null; right: Node | null; constructor(value: number) { this.value = value; this.left = null; this.right = null; } } constructor takes value and sets left and right to null, matching typical binary tree node setup.

3. Final Answer:

   Correct TypeScript class with nullable children and proper constructor -> Option C

4. Quick Check:

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

Hint: Child nodes must allow null to represent empty branches [OK]

Common Mistakes:

• Not initializing left and right to null
• Using undefined instead of null for children
• Missing constructor parameter for value

3. Given the following TypeScript code, 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);
console.log(root.left?.value);

medium

A. 5

B. undefined

C. null

D. 10

Solution

1. Step 1: Analyze node assignments

   root.left is assigned a new Node with value 5, so root.left is not null.

2. Step 2: Understand optional chaining and output

   Using root.left?.value safely accesses value of left node, which is 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   root.left.value = 5 [OK]

Hint: Optional chaining returns value if node exists, else undefined [OK]

Common Mistakes:

• Confusing null and undefined outputs
• Assuming root.left is null by default
• Misreading optional chaining operator

4. Identify the error in this TypeScript binary tree node class and fix it:

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

medium

A. left and right should allow null type (Node | null)

B. Constructor should not assign null to left and right

C. value should be optional in constructor

D. Class should not have a constructor

Solution

1. Step 1: Check property types vs assigned values

   left and right are typed as Node but assigned null, causing type error.

2. Step 2: Fix type declarations

   Change left and right types to Node | null to allow null assignment.

3. Final Answer:

   left and right should allow null type (Node | null) -> Option A

4. Quick Check:

   Nullable children fix type error [OK]

Hint: Child nodes must be nullable if assigned null [OK]

Common Mistakes:

• Not updating types when assigning null
• Removing constructor unnecessarily
• Making value optional without reason

5. How would you modify the binary tree node class to include a parent pointer, and what is the impact on tree traversal?

hard

A. No changes needed; parent pointer is implicit in left and right

B. Add a parent property of type number; no impact on traversal

C. Add a parent property as an array of nodes; allows multiple parents

D. Add a parent property of type Node | null; allows upward traversal from child to parent

Solution

1. Step 1: Understand parent pointer addition

   Adding a parent property of type Node | null stores reference to parent node.

2. Step 2: Impact on traversal

   This allows traversing upwards from child to parent, enabling bidirectional navigation.

3. Final Answer:

   Add a parent property of type Node | null; allows upward traversal from child to parent -> Option D

4. Quick Check:

   Parent pointer enables upward traversal [OK]

Hint: Parent pointer lets you move up the tree easily [OK]

Common Mistakes:

• Using wrong type for parent pointer
• Assuming parent pointer exists by default
• Confusing parent pointer with multiple parents