Binary Tree Node Structure in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a binary tree node with a value property.

DSA Javascript

class TreeNode {
  constructor(val) {
    this.val = [1];
    this.left = null;
    this.right = null;
  }
}

Drag options to blanks, or click blank then click option'

Aval

Bvalue

Cnode

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name like 'value' instead of 'val'.

Forgetting to assign the value to this.val.

The constructor parameter is named 'val', so we assign this.val = val to store the node's value.

2fill in blank

medium

Complete the code to create a new TreeNode with value 5.

DSA Javascript

const node = new TreeNode([1]);

Drag options to blanks, or click blank then click option'

Anull

Bundefined

C'5'

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Passing '5' as a string instead of the number 5.

Passing null or undefined instead of a value.

We pass the number 5 to the TreeNode constructor to create a node with value 5.

3fill in blank

hard

Fix the error in the code to correctly assign the left child of a node.

DSA Javascript

node.[1] = new TreeNode(3);

Drag options to blanks, or click blank then click option'

Aright

Bchild

Cleft

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning to 'right' instead of 'left'.

Using an incorrect property name like 'child' or 'node'.

The left child of a binary tree node is assigned to the 'left' property.

4fill in blank

hard

Fill both blanks to check if a node has a right child and assign it to a variable.

DSA Javascript

if (node.[1] !== [2]) {
  const rightChild = node.right;
}

Drag options to blanks, or click blank then click option'

Aright

Bnull

Cleft

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Checking 'left' instead of 'right'.

Comparing to undefined instead of null.

We check if node.right is not null to confirm the right child exists.

5fill in blank

hard

Fill all three blanks to create a node, assign left and right children with values 1 and 2.

DSA Javascript

const root = new TreeNode([1]);
root.[2] = new TreeNode(1);
root.[3] = new TreeNode(2);

Drag options to blanks, or click blank then click option'

A0

Bleft

Cright

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'root' as a property name instead of 'left' or 'right'.

Assigning children to wrong properties.

We create root with value 0, assign left child to 'left', and right child to 'right'.

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