Binary Tree Node Structure in DSA Javascript

Mental Model

A binary tree node holds a value and links to up to two child nodes, left and right.

Analogy: Think of a family tree where each person can have up to two children, one on the left and one on the right.

    [Node]
    /    \
 [Left] [Right]

Each node points to two children or null if none.

Dry Run Walkthrough

Input: Create a binary tree node with value 10, left child 5, right child 15

Goal: Build a node that connects to two child nodes correctly

Step 1: Create root node with value 10

[10] -> left: null, right: null

Why: Start with the main node that will hold the value

Step 2: Create left child node with value 5

[10] -> left: [5], right: null

Why: Add left child to root to build the left branch

Step 3: Create right child node with value 15

[10] -> left: [5], right: [15]

Why: Add right child to root to build the right branch

Result:

[10] -> left: [5], right: [15]

Annotated Code

DSA Javascript

class TreeNode {
  constructor(value) {
    this.value = value; // store the node's value
    this.left = null;   // pointer to left child
    this.right = null;  // pointer to right child
  }
}

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

// Function to print tree node and its children
function printNode(node) {
  if (!node) return 'null';
  const leftVal = node.left ? node.left.value : 'null';
  const rightVal = node.right ? node.right.value : 'null';
  return `[${node.value}] -> left: [${leftVal}], right: [${rightVal}]`;
}

console.log(printNode(root));

this.value = value; // store the node's value

assign the node's value

this.left = null; // pointer to left child

initialize left child pointer to null

this.right = null; // pointer to right child

initialize right child pointer to null

root.left = new TreeNode(5);

link left child node to root

root.right = new TreeNode(15);

link right child node to root

Output

[10] -> left: [5], right: [15]

Complexity Analysis

Time: O(1) because creating a node and linking children are constant time operations

Space: O(1) for each node created as it stores fixed number of pointers and a value

vs Alternative: Compared to arrays, nodes use pointers to link children, allowing flexible tree shapes without resizing

Edge Cases

Node with no children

left and right pointers remain null, representing leaf node

DSA Javascript

this.left = null;   // pointer to left child

Node with only left child

right pointer stays null, left points to child node

DSA Javascript

this.right = null;  // pointer to right child

Node with only right child

left pointer stays null, right points to child node

DSA Javascript

this.right = null;  // pointer to right child

When to Use This Pattern

When you need to represent hierarchical data with up to two children per item, use the binary tree node structure to link nodes simply.

Common Mistakes

Mistake: Forgetting to initialize left and right pointers to null
Fix: Always set left and right to null in constructor to avoid undefined references

Mistake: Assigning children values directly instead of creating new nodes
Fix: Create new TreeNode instances for children to maintain structure

Summary

It creates a node with a value and two pointers for left and right children.

Use it when building trees where each node can have up to two children.

Remember each node holds its value and links to children or null if none.

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