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

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the maximum element in a BST

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

DSA C

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

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

int main() {
    Node n1 = {10, NULL, NULL};
    Node n2 = {20, NULL, NULL};
    Node n3 = {5, NULL, NULL};
    Node n4 = {15, NULL, NULL};
    n1.right = &n2;
    n1.left = &n3;
    n2.left = &n4;
    int max = findMax(&n1);
    printf("%d\n", max);
    return 0;
}

A 5

B 15

C 20

D 10

💡 Hint

The maximum element in a BST is the rightmost node.

In a BST, the maximum value is found by going as far right as possible. Here, node with value 20 is the rightmost node.

❓ Predict Output

intermediate

Output when BST is empty

What will be the output of the following code when the BST is empty (root is NULL)?

DSA C

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

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

int main() {
    Node* root = NULL;
    int max = findMax(root);
    printf("%d\n", max);
    return 0;
}

A -1

B 0

C Segmentation fault

D Undefined behavior

💡 Hint

Check the base case when root is NULL.

The function returns -1 explicitly when the root is NULL, so the output is -1.

🔧 Debug

advanced

Identify the error in BST maximum element function

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

DSA C

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;
}

int main() {
    Node* root = NULL;
    int max = findMax(root);
    printf("%d\n", max);
    return 0;
}

A Returns -1

B Segmentation fault

C Compilation error

D Prints 0

💡 Hint

What happens if root is NULL and you access root->right?

The function does not check if root is NULL before accessing root->right, causing a segmentation fault.

🧠 Conceptual

advanced

Why does finding maximum in BST take O(h) time?

Why does the function to find the maximum element in a BST take time proportional to the height (h) of the tree?

A Because it traverses the left subtree completely

B Because it visits every node in the tree once

C Because it performs a breadth-first search on all levels

D Because it traverses only the right children from root to the deepest right node

💡 Hint

Think about the path taken to find the maximum in a BST.

The maximum element is found by moving right from the root until no more right child exists, which is at most the height of the tree.

❓ Predict Output

expert

Output of modified BST maximum function with recursion

What is the output of the following recursive function to find the maximum element in a BST?

DSA C

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

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

int main() {
    Node n1 = {8, NULL, NULL};
    Node n2 = {12, NULL, NULL};
    Node n3 = {3, NULL, NULL};
    Node n4 = {10, NULL, NULL};
    n1.left = &n3;
    n1.right = &n2;
    n2.left = &n4;
    int max = findMax(&n1);
    printf("%d\n", max);
    return 0;
}

A 12

B 10

C 8

D -1

💡 Hint

The maximum is the rightmost node's data.

The function recursively goes right until it finds the node with no right child, which is 12 here.

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