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.

// Find minimum value in BST
int findMin(struct Node* root) {
    struct Node* current = root;
    while (current != NULL && current->left != NULL) {
        current = current->left;
    }
    return current->data;
}
    

This code moves left down the tree until it finds the leftmost node, which holds the minimum value.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving down the left child nodes in a loop.
• How many times: Once for each level down the left side until the smallest node is found.

How Execution Grows With Input

As the tree grows, the time depends on how deep the left branch is.

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

Pattern observation: The number of steps grows slowly, roughly with the height of the tree, not the total nodes.

Final Time Complexity

Time Complexity: O(h)

This means the time depends on the height of the tree, which is how many levels it has from top to bottom.

Common Mistake

[X] Wrong: "Finding the minimum always takes time proportional to the total number of nodes (n)."

[OK] Correct: The search only follows the left branch, so it depends on the tree height, not total nodes.

Interview Connect

Understanding how tree height affects search time helps you explain why balanced trees are faster and why this operation is efficient.

Self-Check

"What if the BST is completely unbalanced and looks like a linked list? How would the time complexity change?"

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