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 store this information, you need a basic structure to represent each person and their children.

🎯 Goal: Create a basic binary tree node structure in C++ to represent each person in the family tree with their name and pointers to their left and right children.

📋 What You'll Learn

Define a struct called Node

Include a string variable called name to store the person's name

Include two Node* pointers called left and right for the children

Initialize left and right to nullptr in the constructor

💡 Why This Matters

🌍 Real World

Binary trees are used in many applications like family trees, decision trees, and organizing data for quick search.

💼 Career

Understanding how to create and use binary tree nodes is fundamental for software development roles involving data structures and algorithms.

1

Create the Node struct with a name variable

Define a struct called Node with a public string variable named name.

DSA C++

// Define struct Node with a string variable name
// Your code here

Hint

Use struct Node { string name; }; to define the structure.

2

Add left and right child pointers

Inside the Node struct, add two public pointers named left and right of type Node*.

DSA C++

#include 
using namespace std;

struct Node {
    string name;
    // Add Node* pointers left and right here
    // Your code here
};

Hint

Declare Node* left; and Node* right; inside the struct.

3

Add a constructor to initialize name and children

Add a constructor to Node that takes a string parameter name and initializes the name variable. Also, initialize left and right to nullptr.

DSA C++

#include 
using namespace std;

struct Node {
    string name;
    Node* left;
    Node* right;

    // Add constructor here
    // Your code here
};

Hint

Use an initializer list in the constructor to set name, left, and right.

4

Create a Node instance and print its name

Create a Node object named root with the name "Grandparent". Then print root.name using std::cout.

DSA C++

#include 
#include 
using namespace std;

struct Node {
    string name;
    Node* left;
    Node* right;

    Node(string name) : name(name), left(nullptr), right(nullptr) {}
};

// Create Node root with name "Grandparent" and print root.name
// Your code here

Hint

Use Node root("Grandparent"); and cout << root.name << endl;.

Practice

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

easy

A. A data value and pointers to left and right child nodes

B. Only a data value without any pointers

C. Pointers to parent and sibling nodes only

D. A data value and a pointer to the next node in a list

Solution

1. Step 1: Understand binary tree node components

   A binary tree node stores a data value and has two pointers: one to the left child and one to the right child.

2. Step 2: Compare options with definition

   A data value and pointers to left and right child nodes correctly describes this structure. Other options miss pointers or mention unrelated pointers.

3. Final Answer:

   A data value and pointers to left and right child nodes -> Option A

4. Quick Check:

   Binary tree node = data + left + right pointers [OK]

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

Common Mistakes:

• Thinking nodes have only one pointer
• Confusing with linked list node structure
• Assuming parent pointer is always present

2. Which of the following is the correct syntax to define a binary tree node struct in C++?

easy

A. struct Node { int data; Node* left; Node* right; };

B. struct Node { int data; Node left; Node right; };

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

D. struct Node { int data; Node left*; Node right*; };

Solution

1. Step 1: Check pointer syntax in struct

   In C++, pointers are declared with '*', so left and right children must be pointers: Node* left; Node* right;

2. Step 2: Validate each option

   struct Node { int data; Node* left; Node* right; }; uses correct pointer syntax. struct Node { int data; Node left; Node right; }; uses objects (causes infinite recursion). struct Node { int data; Node& left; Node& right; }; uses references incorrectly. struct Node { int data; Node left*; Node right*; }; has invalid pointer syntax.

3. Final Answer:

   struct Node { int data; Node* left; Node* right; }; -> Option A

4. Quick Check:

   Pointer syntax = Node* left/right [OK]

Hint: Pointers use * after type, not before variable name [OK]

Common Mistakes:

• Using objects instead of pointers for children
• Using references (&) for child nodes
• Incorrect pointer syntax like 'left*'

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

struct Node {
  int data;
  Node* left;
  Node* right;
  Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

int main() {
  Node* root = new Node(10);
  root->left = new Node(5);
  root->right = new Node(15);
  std::cout << root->left->data << " " << root->right->data << std::endl;
  return 0;
}

medium

A. Compilation error

B. 5 15

C. 5 10

D. 10 15

Solution

1. Step 1: Understand node creation and assignments

   Root node has data 10. Left child is assigned a node with data 5. Right child is assigned a node with data 15.

2. Step 2: Trace the output statement

   Output prints root->left->data (5) and root->right->data (15) separated by space.

3. Final Answer:

   5 15 -> Option B

4. Quick Check:

   Left child = 5, Right child = 15 [OK]

Hint: Follow pointer assignments to print child node data [OK]

Common Mistakes:

• Printing root data instead of children
• Confusing left and right child values
• Forgetting to initialize children

4. Identify the error in the following binary tree node struct and fix it:

struct Node {
  int data;
  Node left;
  Node right;
  Node(int val) : data(val) {}
};

medium

A. Remove constructor and use public members only

B. Add default constructor to initialize data

C. Change 'int data;' to 'int* data;'

D. Change 'Node left;' and 'Node right;' to 'Node* left;' and 'Node* right;'

Solution

1. Step 1: Identify the problem with member declarations

   Declaring 'Node left;' and 'Node right;' causes infinite recursion because each Node contains two Nodes inside it.

2. Step 2: Fix by using pointers for children

   Change to 'Node* left;' and 'Node* right;' to hold addresses of child nodes, avoiding infinite size.

3. Final Answer:

   Change 'Node left;' and 'Node right;' to 'Node* left;' and 'Node* right;' -> Option D

4. Quick Check:

   Child nodes must be pointers to avoid infinite recursion [OK]

Hint: Child nodes must be pointers, not objects [OK]

Common Mistakes:

• Using objects for children causing infinite size
• Ignoring pointer usage in tree nodes
• Trying to fix with constructors only

5. You want to create a binary tree node struct that also stores a pointer to its parent node. Which of the following is the correct way to define it?

hard

A. struct Node { int data; Node left; Node right; Node parent; Node(int val) : data(val) {} };

B. struct Node { int data; Node* left; Node* right; int* parent; Node(int val) : data(val), left(nullptr), right(nullptr), parent(nullptr) {} };

C. struct Node { int data; Node* left; Node* right; Node* parent; Node(int val) : data(val), left(nullptr), right(nullptr), parent(nullptr) {} };

D. struct Node { int data; Node* left; Node* right; Node& parent; Node(int val) : data(val), left(nullptr), right(nullptr), parent(*this) {} };

Solution

1. Step 1: Understand parent pointer type

   The parent pointer must be of type 'Node*' to point to another node or nullptr if no parent.

2. Step 2: Check constructor initialization

   Constructor initializes data and sets left, right, and parent pointers to nullptr correctly.

3. Step 3: Validate options

   struct Node { int data; Node* left; Node* right; Node* parent; Node(int val) : data(val), left(nullptr), right(nullptr), parent(nullptr) {} }; correctly uses pointers for children and parent. struct Node { int data; Node left; Node right; Node parent; Node(int val) : data(val) {} }; uses objects causing recursion. struct Node { int data; Node* left; Node* right; int* parent; Node(int val) : data(val), left(nullptr), right(nullptr), parent(nullptr) {} }; wrongly uses 'int*' for parent. struct Node { int data; Node* left; Node* right; Node& parent; Node(int val) : data(val), left(nullptr), right(nullptr), parent(*this) {} }; uses reference incorrectly and initializes parent to self.

4. Final Answer:

   struct Node { int data; Node* left; Node* right; Node* parent; Node(int val) : data(val), left(nullptr), right(nullptr), parent(nullptr) {} }; -> Option C

5. Quick Check:

   Parent pointer = Node* parent [OK]

Hint: Parent pointer must be Node* and initialized to nullptr [OK]

Common Mistakes:

• Using object or reference for parent pointer
• Incorrect pointer type for parent
• Initializing parent pointer incorrectly