BST Property and Why It Matters in DSA C++ - Build from Scratch

BST Property and Why It Matters

📖 Scenario: You are working with a simple phone book application that stores contacts in a Binary Search Tree (BST). Each contact has a unique phone number. To keep the phone book efficient, you need to understand and use the BST property.

🎯 Goal: Build a small BST with three contacts and verify the BST property by checking the values of the nodes. This will help you understand why the BST property is important for fast searching.

📋 What You'll Learn

Create a struct called Node with an integer phoneNumber and two pointers left and right

Create three nodes with phone numbers 5551234, 5555678, and 5550000

Link the nodes to form a BST where 5551234 is the root, 5550000 is the left child, and 5555678 is the right child

Print the phone numbers of the root, left child, and right child to verify the BST structure

💡 Why This Matters

🌍 Real World

Phone books, contact lists, and many databases use BSTs to organize data for quick search and retrieval.

💼 Career

Understanding BSTs is essential for software engineers working on efficient data storage, search algorithms, and system design.

1

Create the Node struct and three nodes

Create a struct called Node with an integer phoneNumber and two pointers left and right. Then create three Node variables called root, leftChild, and rightChild with phone numbers 5551234, 5550000, and 5555678 respectively.

DSA C++

// Define Node struct and create three nodes with given phone numbers
// Your code here

Hint

Define the struct with the three members. Then create the three nodes with the exact phone numbers given.

2

Link the nodes to form a BST

Link the nodes by setting root.left to &leftChild and root.right to &rightChild to form a BST where the root has the left and right children.

DSA C++

#include 

struct Node {
    int phoneNumber;
    Node* left;
    Node* right;
};

int main() {
    Node root = {5551234, nullptr, nullptr};
    Node leftChild = {5550000, nullptr, nullptr};
    Node rightChild = {5555678, nullptr, nullptr};
    // Link the nodes to form the BST
    // Your code here
    return 0;
}

Hint

Assign the addresses of leftChild and rightChild to root.left and root.right respectively.

3

Check the BST property by printing node values

Print the phone numbers of root, root.left, and root.right using std::cout to verify the BST structure.

DSA C++

#include 

struct Node {
    int phoneNumber;
    Node* left;
    Node* right;
};

int main() {
    Node root = {5551234, nullptr, nullptr};
    Node leftChild = {5550000, nullptr, nullptr};
    Node rightChild = {5555678, nullptr, nullptr};

    root.left = &leftChild;
    root.right = &rightChild;

    // Print the phone numbers of root, left child, and right child
    // Your code here
    return 0;
}

Hint

Use std::cout to print the phone numbers. Access the left and right child phone numbers using the arrow operator ->.

4

Run and verify the output

Run the program and verify that the output shows the root phone number as 5551234, the left child as 5550000, and the right child as 5555678.

DSA C++

#include 

struct Node {
    int phoneNumber;
    Node* left;
    Node* right;
};

int main() {
    Node root = {5551234, nullptr, nullptr};
    Node leftChild = {5550000, nullptr, nullptr};
    Node rightChild = {5555678, nullptr, nullptr};

    root.left = &leftChild;
    root.right = &rightChild;

    std::cout phoneNumber phoneNumber << std::endl;

    // Verify output
    // Your code here
    return 0;
}

Hint

Run the program and check the console output matches the expected phone numbers in order.

Practice

1. What is the main property that defines a Binary Search Tree (BST)?

easy

A. For each node, all values in the left subtree are smaller, and all values in the right subtree are larger.

B. Each node has exactly two children.

C. All nodes have the same value.

D. Nodes are arranged in a complete binary tree structure.

Solution

1. Step 1: Understand BST definition

   A BST requires that for every node, values in the left subtree are smaller and values in the right subtree are larger.

2. Step 2: Compare options with BST property

   For each node, all values in the left subtree are smaller, and all values in the right subtree are larger. matches the BST property exactly, while others describe different tree types or incorrect properties.

3. Final Answer:

   For each node, all values in the left subtree are smaller, and all values in the right subtree are larger. -> Option A

4. Quick Check:

   BST property = For each node, all values in the left subtree are smaller, and all values in the right subtree are larger. [OK]

Hint: Remember: left < node < right in BST [OK]

Common Mistakes:

• Thinking BST means every node has two children
• Confusing BST with balanced or complete trees
• Assuming all nodes have the same value

2. Which of the following is the correct C++ syntax to declare a node structure for a BST?

easy

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

Solution

1. Step 1: Check pointer usage in node declaration

   BST nodes must use pointers for left and right children to allow dynamic linking.

2. Step 2: Compare struct and class syntax

   struct Node { int data; Node* left; Node* right; }; correctly uses struct with pointers. class Node { int data; Node* left; Node* right; }; uses class but misses access specifiers (defaults private). Declarations using objects instead of pointers for children are incorrect.

3. Final Answer:

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

4. Quick Check:

   Use pointers for children in BST node [OK]

Hint: Use pointers for child nodes in BST struct [OK]

Common Mistakes:

• Using objects instead of pointers for children
• Forgetting semicolon after struct
• Using class without public access for members

3. Consider the following BST insertion code snippet in C++:

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

Node* insert(Node* root, int val) {
  if (!root) {
    Node* newNode = new Node{val, nullptr, nullptr};
    return newNode;
  }
  if (val < root->data) root->left = insert(root->left, val);
  else root->right = insert(root->right, val);
  return root;
}

int main() {
  Node* root = nullptr;
  root = insert(root, 10);
  root = insert(root, 5);
  root = insert(root, 15);
  root = insert(root, 12);
  root = insert(root, 18);
  return 0;
}

What is the in-order traversal output of the BST after these insertions?

medium

A. [10, 12, 5, 15, 18]

B. [10, 5, 15, 12, 18]

C. [18, 15, 12, 10, 5]

D. [5, 10, 12, 15, 18]

Solution

1. Step 1: Understand insertion order and BST structure

   Inserted values: 10 (root), 5 (left), 15 (right), 12 (left of 15), 18 (right of 15).

2. Step 2: Perform in-order traversal (left, root, right)

   Traverse left subtree: 5, root: 10, left of right subtree: 12, root of right subtree: 15, right of right subtree: 18. So output is [5, 10, 12, 15, 18].

3. Final Answer:

   [5, 10, 12, 15, 18] -> Option D

4. Quick Check:

   In-order traversal of BST = sorted values [OK]

Hint: In-order traversal of BST gives sorted values [OK]

Common Mistakes:

• Confusing pre-order or post-order with in-order
• Ignoring BST insertion rules
• Mixing up left and right subtree traversal

4. The following code attempts to insert a value into a BST but has a bug:

Node* insert(Node* root, int val) {
  if (root == nullptr) {
    Node* newNode = new Node{val, nullptr, nullptr};
    return newNode;
  }
  if (val <= root->data) root->left = insert(root->left, val);
  else root->right = insert(root->right, val);
  return root;
}

What is the main issue with this code regarding BST property?

medium

A. It does not allocate memory for new nodes.

B. Using <= allows duplicate values on the left, which may violate strict BST property.

C. It uses recursion incorrectly causing infinite loops.

D. It does not handle the case when root is null.

Solution

1. Step 1: Analyze the condition for left subtree insertion

   The code uses val <= root->data to decide left subtree insertion, allowing duplicates on the left.

2. Step 2: Understand BST strict property

   Strict BST property requires all left subtree values to be strictly less than the node's value. Allowing duplicates on left can break this.

3. Final Answer:

   Using <= allows duplicate values on the left, which may violate strict BST property. -> Option B

4. Quick Check:

   Strict BST: left < node < right [OK]

Hint: Use < for left subtree to avoid duplicates breaking BST [OK]

Common Mistakes:

• Ignoring duplicate handling in BST
• Thinking recursion causes infinite loops here
• Assuming null root case is missing

5. You have a BST where each node stores an integer. You want to find the kth smallest element efficiently. Which approach best uses the BST property to solve this?

hard

A. Perform an in-order traversal and stop when the kth element is reached.

B. Perform a level-order traversal and count nodes until kth is found.

C. Traverse only the right subtree k times to find the kth smallest.

D. Use pre-order traversal and store all nodes in a list, then sort it.

Solution

1. Step 1: Recall BST in-order traversal property

   In-order traversal visits nodes in ascending order for BST.

2. Step 2: Use in-order traversal to find kth smallest

   By traversing in-order and counting nodes, we can stop at kth node, efficiently finding kth smallest.

3. Final Answer:

   Perform an in-order traversal and stop when the kth element is reached. -> Option A

4. Quick Check:

   In-order traversal = sorted order [OK]

Hint: In-order traversal visits nodes in ascending order [OK]

Common Mistakes:

• Using level-order which is not sorted
• Traversing only right subtree misses smaller values
• Sorting after pre-order wastes time