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

Recall & Review

beginner

What is the minimum element in a Binary Search Tree (BST)?

The minimum element in a BST is the leftmost node because all smaller values are stored to the left.

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 is the minimum.

beginner

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

Because in a BST, left children are always smaller than their parent nodes, so the leftmost node holds the smallest value.

intermediate

What is the time complexity to find the minimum element in a BST?

The time complexity is O(h), where h is the height of the BST. In the worst case (skewed tree), it can be O(n).

beginner

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

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

In a BST, where is the minimum element located?

A Root node

B Rightmost node

C Leftmost node

D Any leaf node

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

What is the first step to find the minimum element in a BST?

A Move to the left child

B Move to the right child

C Check the root value

D Traverse all nodes

You start at the root and move left repeatedly to find the minimum.

If a BST has only one node, what is the minimum element?

A The root node

B No minimum element

C Left child

D Right child

With one node, the root itself is the minimum.

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

A O(n)

B O(log n)

C O(1)

D O(n log n)

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

Which of these is a correct way to find the minimum element in a BST?

A Traverse right children until null

B Traverse both children randomly

C Check only the root node

D Traverse left children until null

Moving left until no left child is the correct method.

Explain how to find the minimum element in a Binary Search Tree and why this method works.

Think about the BST rule for left children.

Write a simple C++ function to find the minimum element in a BST and describe its time complexity.

Focus on traversal and height of tree.

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