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

Time Complexity: Binary Tree Node Structure

O(n)

Understanding Time Complexity

We want to understand how the time to create and link nodes in a binary tree grows as we add more nodes.

How does the work change when the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class TreeNode {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

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

This code creates a simple binary tree node structure and links child nodes to the root.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Creating and linking each node.
• How many times: Once per node created; no loops or recursion here.

How Execution Grows With Input

Each new node requires a fixed amount of work to create and link.

Input Size (n)	Approx. Operations
10	10 node creations and links
100	100 node creations and links
1000	1000 node creations and links

Pattern observation: The work grows directly with the number of nodes added.

Final Time Complexity

Time Complexity: O(n)

This means the time to build the tree grows linearly with the number of nodes.

Common Mistake

[X] Wrong: "Creating a node takes constant time, so the whole tree is always O(1)."

[OK] Correct: While one node is quick, building many nodes adds up, so total time grows with the number of nodes.

Interview Connect

Understanding how node creation scales helps you reason about tree operations and their costs in real problems.

Self-Check

"What if we added a loop to create nodes instead of individual statements? How would the time complexity change?"

Practice

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

easy

A. A value and multiple child nodes in a list

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

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

D. Only a value without any child references

Solution

1. Step 1: Understand binary tree node components

   A binary tree node stores a value and has exactly two child references: left and right.

2. Step 2: Compare options with binary tree 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 left and right 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 JavaScript syntax to define a binary tree node class?

easy

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

B. function Node(value) { this.value = value; this.next = null; }

C. class Node { constructor(value) { this.value = value; this.children = []; } }

D. var Node = (value) => { this.value = value; this.left = null; this.right = null; }

Solution

1. Step 1: Identify correct class syntax for binary tree node

   class Node { constructor(value) { this.value = value; this.left = null; this.right = null; } } uses ES6 class with constructor setting value, left, and right to null.

2. Step 2: Check other options for errors

   function Node(value) { this.value = value; this.next = null; } uses next pointer (linked list), C uses children array (general tree), D uses arrow function incorrectly for constructor.

3. Final Answer:

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

4. Quick Check:

   Class with value, left, right = class Node { constructor(value) { this.value = value; this.left = null; this.right = null; } } [OK]

Hint: Use ES6 class with left and right set to null [OK]

Common Mistakes:

• Using 'next' instead of 'left' and 'right'
• Using arrow function as constructor
• Using children array instead of left/right

3. What will be the output of the following code?

class Node {
  constructor(value) {
    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 + ' -> ' + root.value + ' -> ' + root.right.value);

medium

A. "5 -> 10 -> 15"

B. "10 -> 5 -> 15"

C. "null -> 10 -> null"

D. "15 -> 10 -> 5"

Solution

1. Step 1: Trace node creation and assignments

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

2. Step 2: Evaluate console.log output

   Output concatenates left.value (5), root.value (10), right.value (15) with arrows.

3. Final Answer:

   "5 -> 10 -> 15" -> Option A

4. Quick Check:

   Left 5, root 10, right 15 = "5 -> 10 -> 15" [OK]

Hint: Print left, root, right values in order [OK]

Common Mistakes:

• Mixing order of values in output
• Confusing left and right child values
• Forgetting to access .value property

4. Identify the error in the following binary tree node code snippet:

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

const root = new Node(20);
root.left = new Node(10);
root.right = new Node(30);
root.left.left = 15;
console.log(root.left.left.value);

medium

A. Missing constructor in Node class

B. Using console.log incorrectly

C. Assigning a number directly instead of a Node object to left.left

D. root.right is not assigned properly

Solution

1. Step 1: Analyze assignment to root.left.left

   root.left.left is assigned 15 (a number), not a Node instance.

2. Step 2: Understand why console.log fails

   Accessing .value on a number (15) causes error because 15 has no .value property.

3. Final Answer:

   Assigning a number directly instead of a Node object to left.left -> Option C

4. Quick Check:

   Child nodes must be Node objects, not raw values [OK]

Hint: Child nodes must be Node instances, not raw values [OK]

Common Mistakes:

• Assigning raw values instead of Node objects
• Ignoring error from accessing .value on number
• Assuming constructor is missing

5. Given the following binary tree node class, how would you write a function to count the total number of nodes in the tree starting from the root?

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

Choose the correct function implementation.

hard

A. function countNodes(node) { if (node === null) return null; return 1 + countNodes(node.left) + countNodes(node.right); }

B. function countNodes(node) { if (!node) return 1; return countNodes(node.left) + countNodes(node.right); }

C. function countNodes(node) { return node.value + countNodes(node.left) + countNodes(node.right); }

D. function countNodes(node) { if (!node) return 0; return 1 + countNodes(node.left) + countNodes(node.right); }

Solution

1. Step 1: Understand base case for recursion

   If node is null, count is zero because no node exists there.

2. Step 2: Sum current node plus counts of left and right subtrees

   Return 1 (current node) plus recursive counts of left and right children.

3. Step 3: Check options for correct base case and sum

   function countNodes(node) { if (!node) return 0; return 1 + countNodes(node.left) + countNodes(node.right); } correctly returns 0 for null and sums 1 + left + right counts.

4. Final Answer:

   function countNodes(node) { if (!node) return 0; return 1 + countNodes(node.left) + countNodes(node.right); } -> Option D

5. Quick Check:

   Count nodes = 1 + left count + right count [OK]

Hint: Return 0 for null, else 1 + left + right counts [OK]

Common Mistakes:

• Returning 1 for null node instead of 0
• Adding node.value instead of count 1
• Returning null instead of 0 in base case