BST Find Maximum Element in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find maximum element in a BST

What is the output of the following C++ code that finds the maximum element in a Binary Search Tree?

DSA C++

#include <iostream>

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

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

int main() {
    Node n1 = {10, nullptr, nullptr};
    Node n2 = {20, nullptr, nullptr};
    Node n3 = {5, nullptr, nullptr};
    n1.right = &n2;
    n1.left = &n3;
    int maxVal = findMax(&n1);
    std::cout << maxVal << std::endl;
    return 0;
}

A -1

B 20

C 10

D 5

💡 Hint

The maximum element in a BST is the rightmost node.

The function moves to the right child repeatedly until it reaches the rightmost node, which holds the maximum value 20.

❓ Predict Output

intermediate

Output when BST is empty

What will be the output of the following code when the BST root is nullptr?

DSA C++

#include <iostream>

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

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

int main() {
    Node* root = nullptr;
    int maxVal = findMax(root);
    std::cout << maxVal << std::endl;
    return 0;
}

A -1

B 0

C Segmentation fault

D Undefined behavior

💡 Hint

Check the base case for empty tree in the function.

The function returns -1 immediately if the root is nullptr, so output is -1.

🔧 Debug

advanced

Identify the error in maximum element finder

What error will the following code produce when trying to find the maximum element in a BST?

DSA C++

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

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

A Returns minimum element instead of maximum

B Segmentation fault due to null pointer dereference

C Returns maximum element correctly

D Compilation error due to missing return statement

💡 Hint

Check the direction of traversal inside the loop.

The code incorrectly moves left child inside the loop that checks for right child, so it returns minimum instead of maximum.

🧠 Conceptual

advanced

Maximum element in BST with duplicates

In a BST that allows duplicate values inserted to the right subtree, which node will the findMax function return?

A The root node always

B The leftmost node with the maximum value

C The rightmost node with the maximum value including duplicates

D The minimum value node

💡 Hint

Duplicates go to the right subtree, so maximum is the rightmost node.

Since duplicates are inserted to the right, the maximum value is at the rightmost node including duplicates.

🚀 Application

expert

Find maximum element after BST modification

Given a BST where a new node with value 25 is inserted as the right child of the node with value 20, what will be the output of findMax function?

DSA C++

#include <iostream>

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

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

int main() {
    Node n1 = {10, nullptr, nullptr};
    Node n2 = {20, nullptr, nullptr};
    Node n3 = {5, nullptr, nullptr};
    Node n4 = {25, nullptr, nullptr};
    n1.left = &n3;
    n1.right = &n2;
    n2.right = &n4;
    int maxVal = findMax(&n1);
    std::cout << maxVal << std::endl;
    return 0;
}

A 5

B 20

C 10

D 25

💡 Hint

The maximum is the rightmost node after insertion.

After inserting 25 as right child of 20, the rightmost node is 25, so findMax returns 25.

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