BST Find Minimum Element in DSA C++ - Build from Scratch

BST Find Minimum Element

📖 Scenario: You are working with a Binary Search Tree (BST) that stores numbers. You want to find the smallest number in this tree. This is useful when you want to know the minimum value quickly, like finding the earliest date or lowest price.

🎯 Goal: Build a simple BST with some numbers, then write code to find and print the minimum element in the BST.

📋 What You'll Learn

Create a BST node structure with integer data and left/right pointers

Manually create a BST with these exact values: 15, 10, 20, 8, 12, 17, 25

Write a function called findMin that returns the minimum value in the BST

Print the minimum value found by findMin

💡 Why This Matters

🌍 Real World

Finding minimum values quickly is useful in many applications like scheduling, price comparisons, and searching sorted data.

💼 Career

Understanding BSTs and how to find minimum or maximum elements is a common task in software engineering interviews and real-world coding.

1

Create BST Node Structure and Build Tree

Create a struct called Node with an int data, and two pointers Node* left and Node* right. Then create the BST root node with value 15. Manually add nodes with values 10, 20, 8, 12, 17, and 25 to form the BST.

DSA C++

// Define struct Node
// Create root node with value 15
// Add nodes 10, 20, 8, 12, 17, 25 to build BST
// Your code here

Hint

Remember, in a BST, left child nodes have smaller values and right child nodes have larger values than the parent.

2

Declare the findMin Function

Declare a function called findMin that takes a Node* parameter named root and returns an int. This function will find the minimum value in the BST.

DSA C++

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

Node* root = new Node{15, nullptr, nullptr};
root->left = new Node{10, nullptr, nullptr};
root->right = new Node{20, nullptr, nullptr};
root->left->left = new Node{8, nullptr, nullptr};
root->left->right = new Node{12, nullptr, nullptr};
root->right->left = new Node{17, nullptr, nullptr};
root->right->right = new Node{25, nullptr, nullptr};

int findMin(Node* root);
// Your code here

Hint

Just write the function header for now. The function will return an integer and take a pointer to Node.

3

Implement the findMin Function

Implement the findMin function to find the minimum value in the BST. Use a loop to go left until root->left is nullptr. Return the data of the leftmost node.

DSA C++

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

Node* root = new Node{15, nullptr, nullptr};
root->left = new Node{10, nullptr, nullptr};
root->right = new Node{20, nullptr, nullptr};
root->left->left = new Node{8, nullptr, nullptr};
root->left->right = new Node{12, nullptr, nullptr};
root->right->left = new Node{17, nullptr, nullptr};
root->right->right = new Node{25, nullptr, nullptr};

int findMin(Node* root) {
    // Your code here
}

Hint

Keep moving to the left child until there is no left child. Then return that node's data.

4

Print the Minimum Element

Write a std::cout statement to print the minimum element found by calling findMin(root). Print exactly: Minimum element: X where X is the minimum value.

DSA C++

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

Node* root = new Node{15, nullptr, nullptr};
root->left = new Node{10, nullptr, nullptr};
root->right = new Node{20, nullptr, nullptr};
root->left->left = new Node{8, nullptr, nullptr};
root->left->right = new Node{12, nullptr, nullptr};
root->right->left = new Node{17, nullptr, nullptr};
root->right->right = new Node{25, nullptr, nullptr};

int findMin(Node* root) {
    while (root->left != nullptr) {
        root = root->left;
    }
    return root->data;
}

#include 

int main() {
    // Your code here
}

Hint

Call findMin(root) and print the result with std::cout. Use the exact format: Minimum element: X.

Practice

1. In a Binary Search Tree (BST), where is the minimum element always located?

easy

A. At any leaf node

B. At the leftmost node

C. At the root node

D. At the rightmost node

Solution

1. Step 1: Understand BST property

   In a BST, left child nodes have smaller values than their parents.

2. Step 2: Locate minimum element

   The smallest value is found by going left until no more left child exists.

3. Final Answer:

   At the leftmost node -> Option B

4. Quick Check:

   Minimum = leftmost node [OK]

Hint: Minimum is always the leftmost node in BST [OK]

Common Mistakes:

• Thinking minimum is at root
• Confusing leftmost with rightmost
• Assuming minimum can be any leaf

2. Which of the following C++ code snippets correctly finds the minimum element in a BST node pointer root?

easy

A. while(root->left != nullptr) root = root->left; return root->data;

B. while(root->right != nullptr) root = root->right; return root->data;

C. if(root == nullptr) return INT_MAX; return root->data;

D. return root->data;

Solution

1. Step 1: Check traversal direction

   Minimum is found by moving left until no left child exists.

2. Step 2: Verify code correctness

   while(root->left != nullptr) root = root->left; return root->data; loops left correctly and returns the leftmost node's data.

3. Final Answer:

   while(root->left != nullptr) root = root->left; return root->data; -> Option A

4. Quick Check:

   Traverse left to find minimum [OK]

Hint: Loop left child until null to find minimum [OK]

Common Mistakes:

• Looping right child instead of left
• Returning root data without traversal
• Not checking for nullptr before access

3. What is the output of the following C++ code snippet?

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

Node* root = new Node{10, nullptr, nullptr};
root->left = new Node{5, nullptr, nullptr};
root->right = new Node{15, nullptr, nullptr};
root->left->left = new Node{2, nullptr, nullptr};

Node* temp = root;
while(temp->left != nullptr) {
  temp = temp->left;
}
std::cout << temp->data << std::endl;

medium

A. 10

B. 5

C. 2

D. 15

Solution

1. Step 1: Trace the tree structure

   Root is 10, left child 5, left-left child 2, right child 15.

2. Step 2: Follow the loop to find minimum

   Loop moves from 10 to 5, then to 2, which has no left child.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Leftmost node data = 2 [OK]

Hint: Follow left pointers until null to find minimum [OK]

Common Mistakes:

• Stopping at root or first left child
• Confusing right child as minimum
• Not tracing loop fully

4. Identify the error in the following C++ function to find the minimum element in a BST:

int findMin(Node* root) {
  if(root == nullptr) return -1;
  while(root->right != nullptr) {
    root = root->right;
  }
  return root->data;
}

medium

A. Loops on right child instead of left child

B. Returns -1 when root is nullptr

C. Returns root->data without checking nullptr

D. Function return type should be Node*

Solution

1. Step 1: Analyze loop direction

   Minimum element is found by moving left, not right.

2. Step 2: Identify incorrect traversal

   The code loops on root->right, which finds maximum, not minimum.

3. Final Answer:

   Loops on right child instead of left child -> Option A

4. Quick Check:

   Minimum requires left traversal [OK]

Hint: Minimum is leftmost, not rightmost [OK]

Common Mistakes:

• Using right child loop for minimum
• Ignoring nullptr checks
• Confusing min and max logic

5. Given a BST where some nodes have only right children, how would you modify the standard minimum element search to handle this edge case correctly?

hard

A. Return root data immediately if left child is nullptr

B. Traverse right child until nullptr to find minimum

C. Use recursion to check both left and right children

D. Traverse left child until nullptr; if none, root is minimum

Solution

1. Step 1: Understand edge case

   If no left child exists, root itself is minimum.

2. Step 2: Modify search logic

   Traverse left until nullptr; if none found, root is minimum.

3. Final Answer:

   Traverse left child until nullptr; if none, root is minimum -> Option D

4. Quick Check:

   Minimum = leftmost or root if no left child [OK]

Hint: If no left child, root is minimum [OK]

Common Mistakes:

• Assuming minimum always has left child
• Traversing right child for minimum
• Not handling single-child nodes