BST Find Maximum Element in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the maximum value in the BST.

DSA C

int findMax(struct Node* root) {
    if (root == NULL) {
        return -1; // Tree is empty
    }
    struct Node* current = root;
    while (current->[1] != NULL) {
        current = current->right;
    }
    return current->data;
}

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

Aright

Bleft

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' pointer.

Returning root->data without traversal.

In a BST, the maximum element is the rightmost node. So we keep moving to the right child until it is NULL.

2fill in blank

medium

Complete the recursive function to find the maximum element in the BST.

DSA C

int findMaxRecursive(struct Node* root) {
    if (root == NULL) {
        return -1;
    }
    if (root->[1] == NULL) {
        return root->data;
    }
    return findMaxRecursive(root->right);
}

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

Aleft

Bright

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Checking left child instead of right child.

Not handling NULL root case.

If the right child is NULL, the current node is the maximum. Otherwise, recurse on the right child.

3fill in blank

hard

Fix the error in the iterative maximum finder function.

DSA C

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

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

Aright

Bleft

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using left pointer to find maximum.

Not checking for NULL root.

To find the maximum, we must move to the right child, not left.

4fill in blank

hard

Fill both blanks to complete the recursive maximum finder with base case and recursive call.

DSA C

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

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

A-1

B0

Cdata

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of -1 for empty tree.

Returning left child data instead of node data.

Return -1 for empty tree. Return root->data if no right child.

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 findMaxElement(struct Node* root) {
    if (root == NULL) {
        return [1];
    }
    struct Node* temp = root;
    while (temp->[2] != NULL) {
        temp = temp->[3];
    }
    return temp->data;
}

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

A-1

Bright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using left pointer instead of right.

Not handling empty tree case.

Return -1 if tree is empty. Traverse right child until NULL to find max.

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