BST Find Maximum Element in DSA C - Time & Space Complexity

Time Complexity: BST Find Maximum Element

O(h)

Understanding Time Complexity

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

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Find maximum value in a BST
int findMax(Node* root) {
    if (root == NULL) return -1; // or error value
    Node* current = root;
    while (current->right != NULL) {
        current = current->right;
    }
    return current->data;
}
    

This code moves right in the BST until it finds the rightmost node, which holds the maximum value.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving from one node to its right child in a loop.
• How many times: Up to the height of the tree (number of nodes along the right edge).

How Execution Grows With Input

As the tree grows, the time depends on how tall it is, especially how long the right branch is.

Input Size (n)	Approx. Operations (steps right)
10	Up to 10 (if all nodes are right children)
100	Up to 100
1000	Up to 1000

Pattern observation: The time grows with the height of the tree, which can be as small as log(n) or as large as n.

Final Time Complexity

Time Complexity: O(h) where h is the height of the BST.

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

Common Mistake

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

[OK] Correct: If the tree is very unbalanced, the right branch can be very long, making the search take longer.

Interview Connect

Understanding how tree shape affects search time helps you explain and improve data structure choices in real projects.

Self-Check

"What if the BST was perfectly balanced? How would the time complexity change?"

Practice

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

easy

A. At any leaf node

B. At the leftmost node

C. At the rightmost node

D. At the root node

Solution

1. Step 1: Understand BST property

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

2. Step 2: Locate maximum element

   The maximum element is the rightmost node because it has no right child and is greater than all other nodes.

3. Final Answer:

   At the rightmost node -> Option C

4. Quick Check:

   Maximum = rightmost node [OK]

Hint: Maximum is always the rightmost node in BST [OK]

Common Mistakes:

• Confusing maximum with root node
• Thinking maximum is at leftmost node
• Assuming maximum can be any leaf

2. Which of the following C code snippets correctly finds the maximum element in a BST?

easy

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

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

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

D. return root->data;

Solution

1. Step 1: Analyze code logic

   while(root->right != NULL) root = root->right; return root->data; moves right until no right child exists, which is correct for max in BST.

2. Step 2: Check other options

   while(root->left != NULL) root = root->left; return root->data; moves left (minimum), C returns root data without traversal, D returns root data directly.

3. Final Answer:

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

4. Quick Check:

   Traverse rightmost node for max [OK]

Hint: Traverse right child until NULL to find max [OK]

Common Mistakes:

• Traversing left child instead of right
• Returning root data without traversal
• Not checking for NULL pointers

3. What will be the output of the following code snippet that finds the maximum element in a BST?

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

int findMax(Node* root) {
  while(root->right != NULL) {
    root = root->right;
  }
  return root->data;
}

// BST structure:
//       10
//      /  \
//     5    20
//         /  \
//        15   30

int main() {
  // Assume BST is created as above
  printf("%d", findMax(root));
  return 0;
}

medium

A. 30

B. 20

C. 15

D. 10

Solution

1. Step 1: Understand BST structure

   The rightmost node is 30, which is the maximum element.

2. Step 2: Trace findMax function

   The function moves right from root (10) to 20, then to 30, stops as 30->right is NULL, returns 30.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   Max element = 30 [OK]

Hint: Max is rightmost node: 10 -> 20 -> 30 [OK]

Common Mistakes:

• Returning root data without traversal
• Confusing max with root or left child
• Not following right pointers correctly

4. Identify the error in the following C function that is supposed to find the maximum element in a BST:

int findMax(Node* root) {
  if(root == NULL) return -1;
  while(root->left != NULL) {
    root = root->left;
  }
  return root->data;
}

medium

A. It returns -1 for empty tree instead of INT_MIN

B. It should use recursion instead of loop

C. It does not check if root is NULL before accessing data

D. It traverses left instead of right to find maximum

Solution

1. Step 1: Analyze traversal direction

   The function moves left to find max, but max is at rightmost node in BST.

2. Step 2: Check other issues

   Returning -1 is acceptable for empty tree; checking root NULL is done; recursion is optional.

3. Final Answer:

   It traverses left instead of right to find maximum -> Option D

4. Quick Check:

   Max is rightmost, not leftmost [OK]

Hint: Max element is at rightmost, not leftmost node [OK]

Common Mistakes:

• Confusing left and right child traversal
• Ignoring NULL checks
• Thinking recursion is mandatory

5. Given a BST where some nodes have duplicate values, which approach correctly finds the maximum element considering duplicates are allowed on the right subtree?

int findMax(Node* root) {
  if(root == NULL) return INT_MIN;
  while(root->right != NULL) {
    root = root->right;
  }
  return root->data;
}

What is the output if the BST has nodes with values: 10, 20, 20, 30 (duplicates on right)?

hard

A. 30

B. 20

C. 10

D. Error due to duplicates

Solution

1. Step 1: Understand duplicates in BST

   Duplicates allowed on right means equal values go to right subtree, so max is still rightmost node.

2. Step 2: Trace findMax with duplicates

   Traversal moves right from 10 to 20, then to next 20, then to 30, which is rightmost and max.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   Duplicates on right do not affect max = rightmost node [OK]

Hint: Duplicates on right subtree keep max at rightmost node [OK]

Common Mistakes:

• Assuming duplicates cause errors
• Stopping at first duplicate value
• Confusing duplicates location in BST