BST Find Maximum Element in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum element in a BST by moving to the right child.

DSA C++

int findMax(Node* root) {
    Node* current = root;
    while (current->[1] != nullptr) {
        current = current->right;
    }
    return current->data;
}

Drag options to blanks, or click blank then click option'

Aleft

Bchild

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' to traverse.

Trying to access a non-existent 'child' pointer.

In a BST, the maximum element is found by moving to the right child until there is no right child left.

2fill in blank

medium

Complete the code to return -1 if the BST is empty before finding the maximum element.

DSA C++

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

Drag options to blanks, or click blank then click option'

A0

BNULL

Cnullptr

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or false instead of nullptr.

Using NULL which is legacy in modern C++.

To check if the tree is empty, compare root with nullptr in modern C++.

3fill in blank

hard

Fix the error in the loop condition to correctly traverse the BST to find the maximum element.

DSA C++

int findMax(Node* root) {
    Node* current = root;
    while (current->right [1] nullptr) {
        current = current->right;
    }
    return current->data;
}

Drag options to blanks, or click blank then click option'

A==

B!=

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which stops immediately if right is not nullptr.

Using comparison operators like '<=' or '>=' which are invalid for pointers.

The loop should continue while current->right is not nullptr to move to the rightmost node.

4fill in blank

hard

Fill both blanks to complete the recursive function to find the maximum element in a BST.

DSA C++

int findMax(Node* root) {
    if (root == nullptr) {
        return -1;
    }
    if (root->right == [1]) {
        return root->data;
    }
    return findMax(root->[2]);
}

Drag options to blanks, or click blank then click option'

Anullptr

Broot

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Checking left child instead of right child.

Returning root instead of root->data.

If there is no right child (nullptr), current node is max; else recurse right.

5fill in blank

hard

Fill all three blanks to create a function that finds the maximum element iteratively and handles empty tree.

DSA C++

int findMax(Node* root) {
    if (root == [1]) {
        return -1;
    }
    Node* current = root;
    while (current->[2] != [3]) {
        current = current->right;
    }
    return current->data;
}

Drag options to blanks, or click blank then click option'

Anullptr

Bleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using left instead of right to traverse.

Using 0 or NULL instead of nullptr.

Check if root is nullptr for empty tree, then move right while right child is not nullptr.

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