BST Find Minimum Element in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A BST is a tree where each node has at most two children. The left child's value is less than the parent's value, and the right child's value is greater than the parent's value.

beginner

How do you find the minimum element in a BST?

Start at the root and keep moving to the left child until you reach a node with no left child. That node contains the minimum value.

beginner

Why does moving left in a BST lead to the minimum element?

Because in a BST, left children always have smaller values than their parents, so the leftmost node is the smallest.

intermediate

What is the time complexity of finding the minimum element in a BST?

The time complexity is O(h), where h is the height of the tree. In the best case (balanced tree), it's O(log n), and in the worst case (skewed tree), it's O(n).

intermediate

Show the C code snippet to find the minimum element in a BST.

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

In a BST, where is the minimum element located?

A Leftmost node

B Rightmost node

C Root node

D Any leaf node

The minimum element is always the leftmost node in a BST.

What should you do if the BST is empty when finding the minimum element?

A Return the rightmost node

B Return the root node

C Return NULL or indicate no element

D Insert a new node

If the BST is empty, there is no minimum element, so return NULL.

What is the time complexity to find the minimum element in a balanced BST with n nodes?

A O(n log n)

B O(n)

C O(1)

D O(log n)

In a balanced BST, height is log n, so finding minimum takes O(log n).

Which pointer do you follow to find the minimum element in a BST?

A Left child pointer

B Random pointer

C Parent pointer

D Right child pointer

You keep moving left child pointers to find the minimum.

What happens if you call findMin on a BST node that has no left child?

A It goes to the right child

B It returns that node as minimum

C It returns NULL

D It causes an error

If no left child exists, that node is the minimum.

Explain step-by-step how to find the minimum element in a BST.

Think about the BST property and left children.

Write a simple C function to find the minimum element in a BST and explain how it works.

Use a while loop to traverse left children.

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