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 the BST.

DSA C

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

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

Aparent

Bright

Cleft

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Checking current->right instead of current->left

Returning root->data without traversing

Using recursion unnecessarily

In a BST, the minimum element is found by going as far left as possible. So, we check current->left until it is NULL.

2fill in blank

medium

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

DSA C

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

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

ANULL

Bfalse

Croot

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if root == 0 instead of root == NULL

Not handling empty tree case

Returning 0 as minimum which may be incorrect

If the root is NULL, it means the tree is empty, so we return an error code like -1.

3fill in blank

hard

Fix the error in the code to correctly find the minimum element.

DSA C

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

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

Achild

Bright

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to current->right instead of current->left

Using current->parent which is not always available

Returning root->data without traversal

To find the minimum, we must move left, not right. So current = current->left is correct.

4fill in blank

hard

Fill both blanks to complete the function that finds the minimum node pointer in the BST.

DSA C

struct Node* findMinNode(struct Node* root) {
    struct Node* current = root;
    while (current->[1] != [2]) {
        current = current->left;
    }
    return current;
}

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

Aleft

Bright

CNULL

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Checking current->right instead of current->left

Comparing to root instead of NULL

Returning data instead of node pointer

We check current->left != NULL and move left until no left child remains, then return current node pointer.

5fill in blank

hard

Fill all three blanks to complete the recursive function that finds the minimum value in the BST.

DSA C

int findMinRecursive(struct Node* root) {
    if (root == [1]) {
        return -1; // empty tree
    }
    if (root->[2] == [3]) {
        return root->data;
    }
    return findMinRecursive(root->left);
}

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

ANULL

Bleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Checking root->right instead of root->left

Returning root->data without checking left child

Not handling empty tree case

If root is NULL, tree is empty. If root->left is NULL, root is minimum. Otherwise, recurse left.

Practice

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

easy

A. At the leftmost node

B. At the rightmost node

C. At the root node

D. At any leaf node

Solution

1. Step 1: Understand BST property

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

2. Step 2: Locate minimum element

   Because smaller values go left, the minimum is at the leftmost node.

3. Final Answer:

   At the leftmost node -> Option A

4. Quick Check:

   Minimum element = leftmost node [OK]

Hint: Minimum is always the farthest left 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?

easy

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

B. while(root->left != NULL) root = root->left; return root->data;

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

D. while(root->right != NULL) root = root->right; return root->data;

Solution

1. Step 1: Check loop condition for minimum

   Minimum is found by moving left while left child exists.

2. Step 2: Verify code correctness

   while(root->left != NULL) root = root->left; return root->data; moves left until no left child, then returns data correctly.

3. Final Answer:

   while(root->left != NULL) root = root->left; return root->data; -> Option B

4. Quick Check:

   Traverse left to find minimum [OK]

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

Common Mistakes:

• Traversing right child instead of left
• Returning root data without traversal
• Incorrect loop condition

3. What will be the output of the following C code snippet for a BST?

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

struct Node* findMin(struct Node* root) {
  while(root->left != NULL) {
    root = root->left;
  }
  return root;
}

// Assume BST is:
//       10
//      /  \
//     5    15
//    / 
//   2

int main() {
  struct Node n2 = {2, NULL, NULL};
  struct Node n5 = {5, &n2, NULL};
  struct Node n15 = {15, NULL, NULL};
  struct Node n10 = {10, &n5, &n15};
  struct Node* minNode = findMin(&n10);
  printf("%d", minNode->data);
  return 0;
}

medium

A. 10

B. 5

C. 2

D. 15

Solution

1. Step 1: Trace findMin function

   Function moves left from root 10 to 5, then to 2, stops as 2 has no left child.

2. Step 2: Identify returned node data

   Returned node is 2, so printf prints 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Leftmost node data = 2 [OK]

Hint: Output is leftmost node's data in BST [OK]

Common Mistakes:

• Confusing leftmost with root
• Assuming right child is minimum
• Not following pointer traversal

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

struct Node* findMin(struct Node* root) {
  while(root->left == NULL) {
    root = root->left;
  }
  return root;
}

medium

A. Loop condition should be root->left != NULL

B. Function should return root->right

C. Missing check for root == NULL before loop

D. Should use recursion instead of loop

Solution

1. Step 1: Analyze loop condition

   Loop runs while root->left == NULL, which means it stops if left child exists, so it never moves left.

2. Step 2: Correct loop condition

   It should run while root->left != NULL to move left until no left child.

3. Final Answer:

   Loop condition should be root->left != NULL -> Option A

4. Quick Check:

   Traverse left while left child exists [OK]

Hint: Loop must continue while left child exists [OK]

Common Mistakes:

• Using == NULL instead of != NULL in loop
• Ignoring NULL root check
• Confusing left and right pointers

5. Given a BST where some nodes have only right children, how would you modify the findMin function to safely find the minimum element without causing a runtime error if the tree is empty?

hard

A. Traverse right child instead of left child

B. Return root->data immediately without traversal

C. Use recursion without base case

D. Add a check at start: if(root == NULL) return NULL; then traverse left

Solution

1. Step 1: Handle empty tree case

   Check if root is NULL at start to avoid dereferencing NULL pointer.

2. Step 2: Traverse left children safely

   If root is not NULL, move left while left child exists to find minimum.

3. Final Answer:

   Add a check at start: if(root == NULL) return NULL; then traverse left -> Option D

4. Quick Check:

   Check NULL root before traversal [OK]

Hint: Always check for NULL root before traversal [OK]

Common Mistakes:

• Not checking for empty tree causing crash
• Traversing right child for minimum
• Missing base case in recursion