Binary Tree Node Structure in DSA C# - Build from Scratch

Binary Tree Node Structure

📖 Scenario: Imagine you are building a simple family tree app. Each person can have up to two children. To start, you need to create the basic building block: a node that holds a person's name and links to their children.

🎯 Goal: Create a binary tree node structure in C# that can hold a person's name and references to their left and right children.

📋 What You'll Learn

Create a class called TreeNode

Add a public string variable called Name to store the person's name

Add two public variables called Left and Right of type TreeNode to link to children

Create a constructor that takes a string parameter to set the Name

💡 Why This Matters

🌍 Real World

Binary trees are used in many apps like family trees, decision trees, and organizing data hierarchically.

💼 Career

Understanding how to build and use tree nodes is essential for software developers working with data structures, algorithms, and system design.

1

Create the TreeNode class with Name variable

Create a public class called TreeNode with a public string variable called Name.

DSA C#

// Your code here

Hint

Start by defining a class named TreeNode. Inside it, add a public string variable called Name to hold the person's name.

2

Add Left and Right child references

Inside the TreeNode class, add two public variables called Left and Right of type TreeNode to represent the left and right children.

DSA C#

public class TreeNode
{
    public string Name;
    // Your code here
}

Hint

Think of Left and Right as pointers to the person's children. They should be of the same type as the class itself.

3

Add constructor to set Name

Add a public constructor to the TreeNode class that takes a string parameter called name and sets the Name variable.

DSA C#

public class TreeNode
{
    public string Name;
    public TreeNode Left;
    public TreeNode Right;

    // Your code here
}

Hint

The constructor helps create a new node with a name already set. Use the parameter name to assign the Name variable.

4

Create a TreeNode instance and print Name

Create a TreeNode object called root with the name "Grandparent". Then print the Name of root using Console.WriteLine.

DSA C#

public class TreeNode
{
    public string Name;
    public TreeNode Left;
    public TreeNode Right;

    public TreeNode(string name)
    {
        Name = name;
    }
}

// Your code here

Hint

Create the root node with the name "Grandparent" and print its name to see the output.

Practice

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

easy

A. Only a value without any references

B. A value and a reference to the next node in a list

C. References to parent and sibling nodes only

D. A value and references to left and right child nodes

Solution

1. Step 1: Understand binary tree node components

   A binary tree node stores data and links to its children nodes.

2. Step 2: Identify correct node structure

   It must have a value and two references: one for left child and one for right child.

3. Final Answer:

   A value and references to left and right child nodes -> Option D

4. Quick Check:

   Binary tree node = value + left + right [OK]

Hint: Remember: binary tree nodes have two child links [OK]

Common Mistakes:

• Thinking nodes only store values without links
• Confusing with linked list node structure
• Assuming nodes have parent or sibling references by default

2. Which of the following is the correct C# syntax to declare a binary tree node class with integer value and left/right children?

easy

A. public struct Node { int val; Node left; Node right; }

B. class Node { int val; Node left, right; }

C. public class Node { public int val; public Node left; public Node right; }

D. public class Node { int val; Node left; Node right; void Node() {} }

Solution

1. Step 1: Check access modifiers and field declarations

   Public class with public fields allows access outside the class, correct for node structure.

2. Step 2: Validate syntax correctness

   public class Node { public int val; public Node left; public Node right; } uses public class, public int val, and public Node left/right fields correctly.

3. Final Answer:

   public class Node { public int val; public Node left; public Node right; } -> Option C

4. Quick Check:

   Public class with public fields = public class Node { public int val; public Node left; public Node right; } [OK]

Hint: Use public class and public fields for node structure [OK]

Common Mistakes:

• Missing public keyword causing access issues
• Using struct instead of class for nodes
• Incorrect constructor syntax or missing access modifiers

3. Given the following code, what will be the output when printing the root node's left child's value?

public class Node {
  public int val;
  public Node left;
  public Node right;
  public Node(int v) { val = v; }
}

Node root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);
Console.WriteLine(root.left.val);

medium

A. 5

B. null

C. 15

D. 10

Solution

1. Step 1: Trace node assignments

   Root node value is 10, left child assigned new Node with value 5.

2. Step 2: Identify printed value

   Printing root.left.val outputs 5, the left child's value.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   root.left.val = 5 [OK]

Hint: Print left child's val to get 5 [OK]

Common Mistakes:

• Printing root.val instead of root.left.val
• Confusing left and right child values
• Assuming uninitialized children print null

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

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

Node root = new Node(10);
root.left.val = 5;

medium

A. Cannot assign val to left child before initializing left node

B. Constructor missing return type

C. left and right should be private fields

D. Node class should be a struct

Solution

1. Step 1: Analyze node initialization

   left and right are initialized to null in constructor, so root.left is null initially.

2. Step 2: Check assignment to left.val

   Trying to assign root.left.val = 5 causes null reference error because root.left is null.

3. Final Answer:

   Cannot assign val to left child before initializing left node -> Option A

4. Quick Check:

   Assigning to null child causes error [OK]

Hint: Initialize child node before assigning its value [OK]

Common Mistakes:

• Assuming child nodes auto-initialize
• Ignoring null reference exceptions
• Confusing constructor syntax errors

5. How would you modify the binary tree node class to include a method that returns true if the node is a leaf (has no children)?

hard

A. public bool IsLeaf() { return val == null; }

B. public bool IsLeaf() { return left == null && right == null; }

C. public bool IsLeaf() { return left != null || right != null; }

D. public bool IsLeaf() { return left == right; }

Solution

1. Step 1: Define leaf node condition

   A leaf node has no left or right children, so both must be null.

2. Step 2: Write method to check children

   Return true if left and right are null, false otherwise.

3. Final Answer:

   public bool IsLeaf() { return left == null && right == null; } -> Option B

4. Quick Check:

   Leaf check = left and right null [OK]

Hint: Leaf means no children: left and right null [OK]

Common Mistakes:

• Checking value instead of children
• Using OR instead of AND in condition
• Comparing left and right references directly