BST Find Maximum Element in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A Binary Search Tree is a tree data structure where each node has at most two children. For every node, all nodes in its left subtree have smaller values, and all nodes in its right subtree have larger values.

beginner

In a BST, where is the maximum element located?

The maximum element is located at the rightmost node of the BST because all values to the right are greater.

beginner

What is the basic approach to find the maximum element in a BST?

Start at the root and keep moving to the right child until there is no right child. The last node you reach is the maximum element.

beginner

Why does moving right in a BST lead to the maximum element?

Because in a BST, right children always have values greater than or equal to their parent nodes, so moving right leads to larger or equal values.

intermediate

What is the time complexity of finding the maximum element in a BST?

The time complexity is O(h), where h is the height of the BST. In the worst case (skewed tree), it can be O(n), and in a balanced tree, it is O(log n).

Where do you find the maximum element in a BST?

A Any leaf node

B Leftmost node

C Root node

D Rightmost node

The maximum element is always at the rightmost node in a BST.

What is the first step to find the maximum element in a BST?

A Move to the right child

B Move to the left child

C Check the root value only

D Traverse all nodes

You start at the root and move to the right child repeatedly to find the maximum.

If a BST has only one node, what is the maximum element?

A The root node itself

B No maximum element

C Left child

D Right child

With only one node, that node is both minimum and maximum.

What is the time complexity to find the maximum element in a balanced BST with n nodes?

A O(n)

B O(1)

C O(log n)

D O(n log n)

In a balanced BST, height is log n, so finding max is O(log n).

What happens if you try to move right but the right child is NULL?

A You need to move left

B You found the maximum element

C The tree is empty

D You must restart from root

If there is no right child, the current node is the maximum.

Explain how to find the maximum element in a Binary Search Tree.

Think about the BST property and where larger values are.

Describe the time complexity of finding the maximum element in a BST and why it depends on the tree's shape.

Consider how many right moves you make.

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