Count Total Nodes in Binary Tree in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return 0 when the node is null.

DSA C#

if (root == [1]) return 0;

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

A0

Btrue

Cfalse

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 for a null node.

Using 'true' or 'false' instead of 'null'.

The base case for counting nodes is when the node is null, which means no node exists, so return 0.

2fill in blank

medium

Complete the code to count nodes recursively by adding left and right subtree counts.

DSA C#

return 1 + CountNodes(root.[1]) + CountNodes(root.[2]);

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

Aright

Bleft

Cchild

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same property for both calls.

Using 'child' or 'node' which are not valid properties.

We count the current node (1) plus all nodes in the left and right subtrees. The recursive calls use root.left and root.right.

3fill in blank

hard

Fix the error in the recursive call to count the right subtree.

DSA C#

return 1 + CountNodes(root.left) + CountNodes(root.[1]);

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

Aleft

Bright

Cchild

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' twice causing double counting of left subtree.

Using invalid property names like 'child' or 'node'.

The right subtree is accessed by the 'right' property, so the recursive call must use root.right.

4fill in blank

hard

Fill both blanks to create a method that counts nodes in a binary tree.

DSA C#

public int CountNodes([1] root) {
    if (root == null) return 0;
    return 1 + CountNodes(root.[2]) + CountNodes(root.right);
}

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

ATreeNode

Bleft

CTree

DBinaryTree

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong parameter types like 'Node' or 'Tree'.

Using 'right' instead of 'left' in the recursive call.

The method parameter type is 'TreeNode'. To count left subtree nodes, use 'root.left'.

5fill in blank

hard

Fill all four blanks to complete the binary tree node class and counting method.

DSA C#

public class TreeNode {
    public int val;
    public TreeNode [1];
    public TreeNode [2];
    public TreeNode(int value) {
        val = value;
        [3] = null;
        [4] = null;
    }
}

public int CountNodes(TreeNode root) {
    if (root == null) return 0;
    return 1 + CountNodes(root.left) + CountNodes(root.right);
}

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

Aleft

Bright

Cvalue

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'value' or 'node' as child names.

Not initializing children to null.

The TreeNode class has 'left' and 'right' children. The constructor initializes these to null using the same names.

Practice

1. What does the function to count total nodes in a binary tree do?

easy

A. Counts only the leaf nodes

B. Counts nodes only on the left subtree

C. Counts only the root node

D. Counts all nodes including root, left, and right children

Solution

1. Step 1: Understand the node counting concept

   The function counts every node in the tree, including root and all children.

2. Step 2: Differentiate from partial counts

   Counting only leaves or root ignores other nodes, so total count includes all nodes.

3. Final Answer:

   Counts all nodes including root, left, and right children -> Option D

4. Quick Check:

   Total nodes = all nodes in tree [OK]

Hint: Count every node visited in tree traversal [OK]

Common Mistakes:

• Counting only leaf nodes
• Ignoring root node
• Counting only one subtree

2. Which of the following is the correct C# syntax to define a binary tree node class for counting nodes?

easy

A. public class Node { public int data; public Node left, right; }

B. class Node { int data; Node left, right; void count() {} }

C. public struct Node { int data; Node left; Node right; }

D. class Node { public int data; public Node left; private Node right; }

Solution

1. Step 1: Check class and member access

   public class Node { public int data; public Node left, right; } correctly defines a public class with public members for data and child nodes.

2. Step 2: Verify syntax correctness

   class Node { int data; Node left, right; void count() {} } misses access modifiers and method is irrelevant; C uses struct which is less common; D has private right node which restricts access.

3. Final Answer:

   public class Node { public int data; public Node left, right; } -> Option A

4. Quick Check:

   Correct class with public members [OK]

Hint: Use public class with public node references [OK]

Common Mistakes:

• Missing public keyword
• Using struct instead of class
• Private child nodes blocking access

3. What is the output of the following C# code that counts nodes in a binary tree?

class Node {
  public int data;
  public Node left, right;
  public Node(int val) { data = val; left = right = null; }
}

int CountNodes(Node root) {
  if (root == null) return 0;
  return 1 + CountNodes(root.left) + CountNodes(root.right);
}

// Tree:
//     1
//    / \
//   2   3
//  / 
// 4

Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);

Console.WriteLine(CountNodes(root));

medium

A. 4

B. 3

C. 5

D. 6

Solution

1. Step 1: Count nodes in the tree structure

   The tree has nodes: 1 (root), 2 (left child), 3 (right child), and 4 (left child of 2), total 4 nodes.

2. Step 2: Apply recursive count function

   CountNodes returns 1 + left subtree count + right subtree count, summing to 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Nodes counted = 4 [OK]

Hint: Add 1 for root plus counts of left and right [OK]

Common Mistakes:

• Counting only children, missing root
• Miscounting null nodes
• Adding extra nodes

4. Identify the error in this C# method to count total nodes in a binary tree:

int CountNodes(Node root) {
  if (root == null) return 1;
  return CountNodes(root.left) + CountNodes(root.right);
}

medium

A. Returns 1 for null node instead of 0

B. Does not count the current root node

C. All of the above

D. Missing addition of 1 for current node

Solution

1. Step 1: Check base case return value

   Returning 1 for null node is incorrect; it should be 0 because null means no node.

2. Step 2: Check counting logic for current node

   Function does not add 1 for the current root node, so it misses counting it.

3. Step 3: Combine errors

   Both base case and missing addition cause wrong total count.

4. Final Answer:

   All of the above -> Option C

5. Quick Check:

   Base case and counting errors combined [OK]

Hint: Base case returns 0; add 1 for current node [OK]

Common Mistakes:

• Returning 1 for null node
• Not adding 1 for current node
• Ignoring base case correctness

5. Given a binary tree where some nodes have only one child, how would you modify the counting function to count only nodes that have at least one child?

hard

A. Count only leaf nodes

B. Count nodes where left or right child is not null, recursively

C. Count all nodes including leaves

D. Count nodes with both children null

Solution

1. Step 1: Define condition for counting nodes

   We want to count nodes that have at least one child, so check if left or right child is not null.

2. Step 2: Modify recursive function accordingly

   Recursively count nodes where left or right child exists, ignoring leaf nodes with no children.

3. Step 3: Implement logic

   Return 1 + counts of children if node has child; else 0 for leaf nodes.

4. Final Answer:

   Count nodes where left or right child is not null, recursively -> Option B

5. Quick Check:

   Count nodes with children only [OK]

Hint: Count nodes with at least one child only [OK]

Common Mistakes:

• Counting all nodes including leaves
• Counting only leaves
• Counting nodes with no children