BST Find Minimum Element in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the minimum value in a BST by going left.

DSA C++

int findMin(Node* root) {
    while (root->[1] != nullptr) {
        root = root->left;
    }
    return root->data;
}

Drag options to blanks, or click blank then click option'

Aleft

Bparent

Cright

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' to find the minimum.

Trying to access a non-existent 'parent' pointer.

In a BST, the minimum element is found by continuously moving to the left child until there is no left child.

2fill in blank

medium

Complete the code to handle the case when the BST is empty.

DSA C++

int findMin(Node* root) {
    if (root == [1]) {
        return -1; // or some error code
    }
    while (root->left != nullptr) {
        root = root->left;
    }
    return root->data;
}

Drag options to blanks, or click blank then click option'

Anullptr

B0

CNULL

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or NULL instead of nullptr in modern C++.

Not checking for empty tree before accessing nodes.

We check if the root pointer is nullptr to know if the tree is empty.

3fill in blank

hard

Fix the error in the recursive version of findMin function.

DSA C++

int findMin(Node* root) {
    if (root == nullptr) {
        return -1;
    }
    if (root->[1] == nullptr) {
        return root->data;
    }
    return findMin(root->left);
}

Drag options to blanks, or click blank then click option'

Aright

Bparent

Cchild

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Checking right child instead of left child.

Using an incorrect pointer name.

The base case checks if the left child is nullptr to find the minimum node.

4fill in blank

hard

Fill both blanks to complete the iterative findMin function with a for loop.

DSA C++

int findMin(Node* root) {
    for (; root != [1]; root = root->[2]) {
        if (root->left == nullptr) {
            break;
        }
    }
    return root->data;
}

Drag options to blanks, or click blank then click option'

Anullptr

BNULL

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using NULL instead of nullptr.

Moving to the right child instead of left.

The loop continues until root->left is nullptr, moving left each time to find the minimum.

5fill in blank

hard

Fill all three blanks to create a recursive function that returns the minimum value in a BST.

DSA C++

int findMin(Node* root) {
    if (root == [1]) {
        return -1;
    }
    if (root->[2] == [3]) {
        return root->data;
    }
    return findMin(root->left);
}

Drag options to blanks, or click blank then click option'

Anullptr

Bleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Checking right child instead of left child.

Not checking for empty tree first.

Check if root is nullptr first, then if left child is nullptr to find minimum recursively.

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