BST Find Minimum Element in DSA C++ - Time & Space Complexity

Time Complexity: BST Find Minimum Element

O(h)

Understanding Time Complexity

We want to understand how long it takes to find the smallest value in a Binary Search Tree (BST).

How does the time needed change as the tree grows bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the minimum element by going left until no more left child exists.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving from a node to its left child repeatedly.
• How many times: At most once per level down the left side of the tree.

How Execution Grows With Input

As the tree grows, the time depends on the height of the leftmost path.

Input Size (n)	Approx. Operations
10	Up to 4 steps (if tree is balanced)
100	Up to 7 steps (balanced tree height)
1000	Up to 10 steps (balanced tree height)

Pattern observation: In a balanced tree, steps grow slowly with size (logarithmically). In a skewed tree, steps can grow linearly.

Final Time Complexity

Time Complexity: O(h) where h is the tree height

This means the time depends on how tall the tree is, not just how many nodes it has.

Common Mistake

[X] Wrong: "Finding minimum always takes the same time regardless of tree shape."

[OK] Correct: If the tree is very unbalanced (like a linked list), it takes longer because you must go through many nodes.

Interview Connect

Understanding how tree shape affects search time is a key skill for working with trees and helps you explain your code clearly in interviews.

Self-Check

"What if the tree was balanced using a self-balancing method? How would the time complexity change?"

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