BST Find Minimum Element in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find minimum element in a BST - Output?

What is the output of the following C# code that finds the minimum element in a Binary Search Tree (BST)?

DSA C#

using System;

class Node {
    public int data;
    public Node left, right;
    public Node(int item) {
        data = item;
        left = right = null;
    }
}

class BST {
    public Node root;

    public int FindMin(Node node) {
        if (node == null) throw new InvalidOperationException("Tree is empty");
        while (node.left != null) {
            node = node.left;
        }
        return node.data;
    }
}

class Program {
    static void Main() {
        BST tree = new BST();
        tree.root = new Node(20);
        tree.root.left = new Node(10);
        tree.root.right = new Node(30);
        tree.root.left.left = new Node(5);
        tree.root.left.right = new Node(15);
        Console.WriteLine(tree.FindMin(tree.root));
    }
}

A 20

B 5

C 15

D 10

💡 Hint

The minimum element in a BST is the leftmost node.

The code traverses the left children starting from the root until it reaches the leftmost node, which holds the minimum value 5.

❓ Predict Output

intermediate

What error occurs when finding minimum in an empty BST?

What error does the following C# code throw when trying to find the minimum element in an empty BST?

DSA C#

using System;

class Node {
    public int data;
    public Node left, right;
    public Node(int item) {
        data = item;
        left = right = null;
    }
}

class BST {
    public Node root;

    public int FindMin(Node node) {
        if (node == null) throw new InvalidOperationException("Tree is empty");
        while (node.left != null) {
            node = node.left;
        }
        return node.data;
    }
}

class Program {
    static void Main() {
        BST tree = new BST();
        Console.WriteLine(tree.FindMin(tree.root));
    }
}

A NullReferenceException

B ArgumentNullException

C InvalidOperationException

D No error, outputs 0

💡 Hint

Check the explicit exception thrown when node is null.

The code explicitly throws InvalidOperationException with message "Tree is empty" when the root node is null.

🔧 Debug

advanced

Why does this FindMin method cause an infinite loop?

Identify the reason for the infinite loop in the following FindMin method for a BST.

DSA C#

public int FindMin(Node node) {
    while (node.left != null) {
        node = node.right;
    }
    return node.data;
}

A It moves right instead of left, so it never reaches null

B It should check node.right instead of node.left in the while condition

C It returns node.data before the loop ends causing infinite recursion

D It does not handle empty tree (null node) causing infinite loop

💡 Hint

The loop condition checks left child but moves to right child inside the loop.

The loop condition checks if node.left is not null but inside the loop it moves to node.right, so node.left never changes and the loop never ends.

🧠 Conceptual

advanced

Minimum element in BST after insertion

Given a BST with minimum element 3, what will be the minimum element after inserting 1 and 4?

A 1

B 3

C 4

D Cannot determine without tree structure

💡 Hint

In BST, smaller values go to the left subtree.

Inserting 1, which is smaller than current minimum 3, will update the minimum element to 1.

🚀 Application

expert

Find minimum element in BST with duplicates allowed

In a BST that allows duplicate values inserted to the right subtree, what is the minimum element after inserting these values in order: 10, 5, 5, 3, 7?

A 5

B 10

C 7

D 3

💡 Hint

Duplicates go to the right subtree, minimum is still the leftmost node.

Even with duplicates inserted to the right, the minimum element is the leftmost node which is 3.

Practice

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

easy

A. At the rightmost node

B. At the root node

C. At the leftmost node

D. At any leaf 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 as far left as possible from the root.

3. Final Answer:

   At the leftmost node -> Option C

4. Quick Check:

   Minimum element = leftmost node [OK]

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

Common Mistakes:

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

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

easy

A. while (root.Left != null) { root = root.Left; } return root.Value;

B. while (root.Right != null) { root = root.Right; } return root.Value;

C. return root.Value;

D. while (root != null) { root = root.Left; } return root.Value;

Solution

1. Step 1: Check loop condition for minimum

   Minimum is found by moving left while left child exists.

2. Step 2: Validate code correctness

   while (root.Left != null) { root = root.Left; } return root.Value; moves left until null, then returns last node's value correctly.

3. Final Answer:

   while (root.Left != null) { root = root.Left; } return root.Value; -> Option A

4. Quick Check:

   Traverse left until null, then return value [OK]

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

Common Mistakes:

• Traversing right child instead of left
• Returning root value without traversal
• Looping until root is null causing error

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

class Node {
  public int Value;
  public Node Left, Right;
  public Node(int val) { Value = val; }
}

Node root = new Node(10);
root.Left = new Node(5);
root.Right = new Node(15);
root.Left.Left = new Node(2);
root.Left.Right = new Node(7);

Node current = root;
while (current.Left != null) {
  current = current.Left;
}
Console.WriteLine(current.Value);

medium

A. 7

B. 2

C. 5

D. 10

Solution

1. Step 1: Trace the tree structure

   Root is 10, left child 5, left-left child 2, left-right child 7.

2. Step 2: Follow the loop to find minimum

   Loop moves from 10 to 5, then to 2 (leftmost node). 2 has no left child, loop stops.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Leftmost node value = 2 [OK]

Hint: Leftmost node value is printed after loop [OK]

Common Mistakes:

• Printing root value instead of leftmost
• Confusing left-right child as minimum
• Stopping loop too early

4. Identify the error in the following C# method to find the minimum element in a BST:

int FindMin(Node root) {
  while (root != null) {
    root = root.Left;
  }
  return root.Value;
}

medium

A. The method should check root.Right instead of root.Left

B. The method should return root.Left.Value instead of root.Value

C. The method should use recursion instead of a loop

D. The loop condition should be root.Left != null, not root != null

Solution

1. Step 1: Analyze loop condition

   Loop runs while root is not null, but root becomes null after last left move.

2. Step 2: Identify cause of error

   After loop, root is null, so accessing root.Value causes error. Condition should be root.Left != null to stop at last node.

3. Final Answer:

   The loop condition should be root.Left != null, not root != null -> Option D

4. Quick Check:

   Stop before null to access valid node [OK]

Hint: Loop until root.Left is null, not root itself [OK]

Common Mistakes:

• Looping until root is null causing null reference
• Returning value after root becomes null
• Checking right child instead of left

5. Given a BST where some nodes have duplicate values, which approach correctly finds the minimum element considering duplicates are allowed only on the right subtree?

class Node {
  public int Value;
  public Node Left, Right;
  public Node(int val) { Value = val; }
}

// Duplicates go to right subtree

Which code snippet correctly finds the minimum element?

hard

A. while (root.Left != null) { root = root.Left; } return root.Value;

B. while (root.Right != null) { root = root.Right; } return root.Value;

C. return root.Value; // root is minimum due to duplicates on right

D. while (root.Left != null && root.Left.Value < root.Value) { root = root.Left; } return root.Value;

Solution

1. Step 1: Understand duplicate placement rule

   Duplicates go to right subtree, so minimum is still in leftmost path.

2. Step 2: Confirm traversal for minimum

   Traverse left child nodes until no more left child exists to find minimum.

3. Final Answer:

   while (root.Left != null) { root = root.Left; } return root.Value; -> Option A

4. Quick Check:

   Minimum is leftmost node even with duplicates on right [OK]

Hint: Duplicates on right don't affect leftmost minimum [OK]

Common Mistakes:

• Checking right subtree for minimum
• Assuming root is minimum without traversal
• Adding extra conditions unnecessarily