Why BST Over Plain Binary Tree in DSA C++ - See It Work

Why BST Over Plain Binary Tree

📖 Scenario: Imagine you have a collection of numbers and you want to find, add, or check if a number exists quickly. You can organize these numbers in a tree structure. A plain binary tree just puts numbers anywhere, but a Binary Search Tree (BST) keeps them in order. This project will help you see why BST is better for searching.

🎯 Goal: You will build a simple binary tree and a binary search tree with the same numbers. Then, you will search for a number in both trees and see the difference in how they work.

📋 What You'll Learn

Create a plain binary tree with specific numbers

Create a binary search tree with the same numbers

Write a function to search a number in the plain binary tree

Write a function to search a number in the BST

Print the search results for both trees

💡 Why This Matters

🌍 Real World

Organizing data so you can find things quickly is important in many apps like phone contacts or product catalogs.

💼 Career

Understanding BST helps in software development roles where efficient data searching and storage is needed.

1

Create a plain binary tree with nodes 10, 5, 15

Create a struct called Node with an int data, and two pointers left and right. Then create three nodes with values 10, 5, and 15. Connect them so that 10 is the root, 5 is the left child, and 15 is the right child.

DSA C++

// Define struct Node
// Create nodes root, leftChild, rightChild
// Connect nodes to form the tree

Hint

Think of Node as a box holding a number and two pointers to other boxes.

2

Create a function to search a number in the plain binary tree

Add a function called searchPlain that takes a Node* and an int key. It should return true if the key is found anywhere in the tree by checking the current node and recursively searching left and right children.

DSA C++

#include 

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

bool searchPlain(Node* root, int key) {
    // Your code here
}

int main() {
    Node* root = new Node{10, nullptr, nullptr};
    Node* leftChild = new Node{5, nullptr, nullptr};
    Node* rightChild = new Node{15, nullptr, nullptr};

    root->left = leftChild;
    root->right = rightChild;

    return 0;
}

Hint

Check if the current node matches the key. If not, search left and right children.

3

Create a function to search a number in the binary search tree (BST)

Add a function called searchBST that takes a Node* and an int key. It should return true if the key is found by using the BST property: if key is less than current node's data, search left; if greater, search right; if equal, return true.

DSA C++

#include 

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

bool searchPlain(Node* root, int key) {
    if (root == nullptr) {
        return false;
    }
    if (root->data == key) {
        return true;
    }
    return searchPlain(root->left, key) || searchPlain(root->right, key);
}

bool searchBST(Node* root, int key) {
    // Your code here
}

int main() {
    Node* root = new Node{10, nullptr, nullptr};
    Node* leftChild = new Node{5, nullptr, nullptr};
    Node* rightChild = new Node{15, nullptr, nullptr};

    root->left = leftChild;
    root->right = rightChild;

    return 0;
}

Hint

Use the rule: go left if key is smaller, right if bigger, stop if equal.

4

Print search results for key 15 in both trees

In main(), call searchPlain(root, 15) and searchBST(root, 15). Print "Found in plain tree" if searchPlain returns true, else "Not found in plain tree". Similarly, print "Found in BST" or "Not found in BST" based on searchBST result.

DSA C++

#include 

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

bool searchPlain(Node* root, int key) {
    if (root == nullptr) {
        return false;
    }
    if (root->data == key) {
        return true;
    }
    return searchPlain(root->left, key) || searchPlain(root->right, key);
}

bool searchBST(Node* root, int key) {
    if (root == nullptr) {
        return false;
    }
    if (root->data == key) {
        return true;
    } else if (key data) {
        return searchBST(root->left, key);
    } else {
        return searchBST(root->right, key);
    }
}

int main() {
    Node* root = new Node{10, nullptr, nullptr};
    Node* leftChild = new Node{5, nullptr, nullptr};
    Node* rightChild = new Node{15, nullptr, nullptr};

    root->left = leftChild;
    root->right = rightChild;

    // Call searchPlain and searchBST for key 15
    // Print results accordingly
}

Hint

Use if statements to check the return values and print the messages.

Practice

1. Why is a Binary Search Tree (BST) preferred over a plain Binary Tree for searching operations?

easy

A. Because BST keeps elements in sorted order allowing faster search

B. Because BST uses less memory than a plain Binary Tree

C. Because BST nodes can have more than two children

D. Because BST does not require pointers to child nodes

Solution

1. Step 1: Understand the structure of BST

   A BST stores elements so that left child nodes are smaller and right child nodes are larger than the parent node, maintaining sorted order.

2. Step 2: Compare search efficiency

   This sorted structure allows searching by comparing values and moving left or right, reducing search time to O(log n) on average, unlike a plain Binary Tree which has no order and requires O(n) search.

3. Final Answer:

   Because BST keeps elements in sorted order allowing faster search -> Option A

4. Quick Check:

   BST sorted order = faster search [OK]

Hint: BST sorts nodes for quick search, plain tree does not [OK]

Common Mistakes:

• Thinking BST uses less memory than plain tree
• Confusing number of children allowed in BST
• Assuming BST nodes don't have child pointers

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

easy

A. struct Node { int data; Node* parent; };

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

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

D. struct Node { int data; Node* next; };

Solution

1. Step 1: Identify correct pointer usage for children

   A BST node must have pointers to left and right child nodes, so Node* left and Node* right are needed.

2. Step 2: Check each option

   struct Node { int data; Node* left; Node* right; }; correctly declares pointers to left and right children. struct Node { int data; Node left; Node right; }; incorrectly uses Node left and right (not pointers). struct Node { int data; Node* parent; }; only has parent pointer, missing children. struct Node { int data; Node* next; }; uses next pointer which is for linked lists, not BST.

3. Final Answer:

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

4. Quick Check:

   BST node needs left and right pointers [OK]

Hint: BST nodes have left and right child pointers [OK]

Common Mistakes:

• Using objects instead of pointers for children
• Confusing parent pointer with child pointers
• Using 'next' pointer like in linked lists

3. Consider the following code inserting values into a BST:

#include <iostream>

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

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

void inorder(Node* node) {
  if (!node) return;
  inorder(node->left);
  std::cout << node->data << " ";
  inorder(node->right);
}

int main() {
  Node* root = nullptr;
  root = insert(root, 10);
  root = insert(root, 5);
  root = insert(root, 15);
  root = insert(root, 7);
  inorder(root);
  return 0;
}

What is the output of the inorder traversal?

medium

A. 10 5 7 15

B. 15 10 7 5

C. 5 7 10 15

D. 7 5 10 15

Solution

1. Step 1: Understand insertion order and BST property

   Values inserted are 10 (root), 5 (left of 10), 15 (right of 10), 7 (right of 5). BST property keeps left < parent < right.

2. Step 2: Perform inorder traversal

   Inorder traversal visits left subtree, node, then right subtree. So nodes visited in order: 5, 7, 10, 15.

3. Final Answer:

   5 7 10 15 -> Option C

4. Quick Check:

   Inorder traversal of BST = sorted order [OK]

Hint: Inorder traversal of BST prints sorted values [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Misplacing 7 as left child of 10
• Ignoring BST insertion rules

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

What is the bug and how to fix it?

medium

A. The recursive calls do not assign returned nodes to root's children; fix by root->left = insert(root->left, val) and root->right = insert(root->right, val)

B. The base case is wrong; should check if val == root->data

C. The function should return void instead of Node*

D. The comparison should be val < root->data instead of val <= root->data

Solution

1. Step 1: Identify missing assignment in recursion

   The recursive calls insert(root->left, val) and insert(root->right, val) do not update root's left or right pointers with the returned new nodes.

2. Step 2: Fix by assigning returned nodes

   Assign the recursive call results back: root->left = insert(root->left, val); root->right = insert(root->right, val); so the tree links update correctly.

3. Final Answer:

   The recursive calls do not assign returned nodes to root's children; fix by root->left = insert(root->left, val) and root->right = insert(root->right, val) -> Option A

4. Quick Check:

   Assign recursive insert results to children [OK]

Hint: Always assign recursive insert results to child pointers [OK]

Common Mistakes:

• Ignoring returned node from recursion
• Changing base case unnecessarily
• Changing comparison operator without reason

5. You have a plain Binary Tree and a BST both containing the same 7 unique values. You want to find if a value exists in the tree. Which of the following statements best explains why searching in the BST is generally faster than in the plain Binary Tree?

hard

A. BST uses hashing internally to find values instantly

B. BST stores values in a linked list internally, making search O(1)

C. Plain Binary Tree nodes have no pointers, so search is impossible

D. BST allows skipping half of the nodes at each step due to ordering, reducing search time to O(log n) on average

Solution

1. Step 1: Understand search in BST vs plain Binary Tree

   BST maintains sorted order so at each node, you decide to go left or right, effectively halving the search space each step.

2. Step 2: Compare time complexities

   This halving leads to average search time O(log n). Plain Binary Tree has no order, so search may check every node, O(n) time.

3. Final Answer:

   BST allows skipping half of the nodes at each step due to ordering, reducing search time to O(log n) on average -> Option D

4. Quick Check:

   BST halves search space each step [OK]

Hint: BST halves search space; plain tree searches all nodes [OK]

Common Mistakes:

• Thinking BST uses linked lists internally
• Believing plain Binary Tree has no pointers
• Assuming BST uses hashing