Binary Tree Node Structure in DSA Typescript - Time & Space Complexity

Time Complexity: Binary Tree Node Structure

O(1)

Understanding Time Complexity

We want to understand how the time to create or access nodes in a binary tree grows as the tree gets bigger.

How does the number of steps change when we add more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

This code defines a single node in a binary tree with a value and two child pointers.

Identify Repeating Operations

Since this snippet only creates one node, there are no loops or recursion here.

• Primary operation: Creating one node with fixed steps.
• How many times: Exactly once per node creation.

How Execution Grows With Input

Each node creation takes the same small number of steps, no matter the tree size.

Input Size (n)	Approx. Operations
10	10 node creations x constant steps = 10 x c
100	100 x c
1000	1000 x c

Pattern observation: The total work grows linearly if we create many nodes one by one.

Final Time Complexity

Time Complexity: O(1) per node creation

This means creating a single node always takes the same small amount of time, no matter what.

Common Mistake

[X] Wrong: "Creating a node takes longer as the tree grows bigger."

[OK] Correct: Each node is created independently with fixed steps, so size of the tree does not affect the time to create one node.

Interview Connect

Understanding that node creation is a constant-time operation helps you focus on the parts of tree algorithms where time grows with input, like traversals or searches.

Self-Check

"What if the constructor also initialized child nodes recursively? How would the time complexity change?"

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