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 biggest number in a Binary Search Tree (BST).

How does the time needed change when the tree grows 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 == nullptr) return -1; // empty tree
    Node* current = root;
    while (current->right != nullptr) {
        current = current->right;
    }
    return current->data;
}
    

This code moves right in the tree until it finds the largest value, which is the rightmost node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving from one node to its right child.
• How many times: Up to the height of the tree (number of right edges).

How Execution Grows With Input

As the tree grows taller, the number of steps to reach the rightmost node grows roughly with the tree height.

Input Size (n)	Approx. Operations (steps to rightmost node)
10	Up to 10 (if tree is a straight line)
100	Up to 100 (worst case)
1000	Up to 1000 (worst case)

Pattern observation: The time grows linearly with the height of the tree, which can be as big as the number of nodes in the worst case.

Final Time Complexity

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

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

Common Mistake

[X] Wrong: "Finding the max in a BST always takes the same time no matter the tree size."

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

Interview Connect

Knowing how tree shape affects search time helps you explain and improve data structures in real projects.

Self-Check

"What if the BST was 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 the leftmost node

B. At any leaf 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 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 D

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++ function signatures correctly finds the maximum element in a BST node pointer root?

easy

A. int findMax(Node* root) { return root->data; }

B. int findMax(Node* root) { while(root->left) root = root->left; return root->data; }

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

D. int findMax(Node* root) { while(root->parent) root = root->parent; return root->data; }

Solution

1. Step 1: Check traversal direction for maximum

   Maximum is found by moving right until no right child exists.

2. Step 2: Verify code correctness

   int findMax(Node* root) { while(root->right) root = root->right; return root->data; } correctly loops while root->right exists and returns root->data.

3. Final Answer:

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

4. Quick Check:

   Traverse right child to find max [OK]

Hint: Traverse right child pointer until null [OK]

Common Mistakes:

• Traversing left child instead of right
• Returning root data without traversal
• Using parent pointer which may not exist

3. What will be the output of the following C++ code snippet?

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

Node* root = new Node{10, nullptr, nullptr};
root->right = new Node{20, nullptr, nullptr};
root->right->right = new Node{30, nullptr, nullptr};

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

std::cout << findMax(root) << std::endl;

medium

A. 30

B. 20

C. 10

D. Runtime error

Solution

1. Step 1: Trace the BST structure

   Root is 10, right child 20, right child's right child 30.

2. Step 2: Follow findMax function

   Loop moves from 10 to 20, then 20 to 30, then stops as 30 has no right child.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   Rightmost node data = 30 [OK]

Hint: Rightmost node data is maximum [OK]

Common Mistakes:

• Returning root data without traversal
• Confusing left and right pointers
• Assuming runtime error due to nullptr

4. Identify the bug in the following C++ function to find the maximum element in a BST:

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

medium

A. Should traverse right child, not left child

B. Should return 0 instead of -1 for nullptr

C. Missing check for root->right in while loop

D. Function should be void, not int

Solution

1. Step 1: Analyze traversal direction

   The function incorrectly traverses left child to find maximum, but maximum is at rightmost node.

2. Step 2: Correct traversal logic

   Change while(root->left) to while(root->right) to find maximum correctly.

3. Final Answer:

   Should traverse right child, not left child -> Option A

4. Quick Check:

   Maximum = rightmost node, not leftmost [OK]

Hint: Maximum is rightmost, so traverse right pointer [OK]

Common Mistakes:

• Traversing left child instead of right
• Ignoring nullptr check
• Returning wrong default value

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?

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

// Function to find max element
int findMax(Node* root) {
  // Your code here
}

hard

A. Traverse right while root->right exists and root->right->data > root->data

B. Traverse right while root->right exists only

C. Traverse left while root->left exists and root->left->data > root->data

D. Return root->data without traversal

Solution

1. Step 1: Understand duplicates in BST

   Even with duplicates allowed on the right subtree, BST property ensures right subtree values are >= parent, so maximum is always the rightmost node.

2. Step 2: Correct traversal for maximum

   Traverse right while root->right exists only. No data comparison needed due to BST invariants.

3. Final Answer:

   Traverse right while root->right exists only -> Option B

4. Quick Check:

   Traverse right child pointer until null [OK]

Hint: Traverse right child pointer until null [OK]

Common Mistakes:

• Using strict > check which stops early on duplicates
• Traversing left instead of right
• Returning root data without traversal