BST Find Maximum Element in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the maximum value in a BST by moving right.

DSA Typescript

function findMax(root: TreeNode | null): number | null {
  if (root === null) return null;
  let current = root;
  while (current.[1] !== null) {
    current = current.right;
  }
  return current.val;
}

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

Aleft

Bparent

Cright

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

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

Returning root value without traversal.

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

2fill in blank

medium

Complete the code to recursively find the maximum value in a BST.

DSA Typescript

function findMaxRecursive(root: TreeNode | null): number | null {
  if (root === null) return null;
  if (root.[1] === null) return root.val;
  return findMaxRecursive(root.right);
}

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

Aright

Bparent

Cleft

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Checking left child instead of right.

Not handling null root case.

The recursion stops when there is no right child, meaning the current node is the maximum.

3fill in blank

hard

Fix the error in the code to correctly find the maximum value in a BST iteratively.

DSA Typescript

function findMaxIterative(root: TreeNode | null): number | null {
  if (root === null) return null;
  let current = root;
  while (current.[1] !== null) {
    current = current.[1];
  }
  return current.val;
}

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

Achild

Bleft

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Moving left instead of right in the loop.

Returning wrong node value.

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

4fill in blank

hard

Fill both blanks to create a function that finds the maximum value in a BST using recursion.

DSA Typescript

function findMax(root: TreeNode | null): number | null {
  if (root === null) return null;
  if (root.[1] === null) return root.val;
  return findMax(root.[2]);
}

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

Aright

Bleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using left child in conditions.

Not returning the recursive call properly.

The function checks if the right child is null to stop recursion, otherwise it continues to the right child.

5fill in blank

hard

Fill all three blanks to implement an iterative function that finds the maximum value in a BST.

DSA Typescript

function findMax(root: TreeNode | null): number | null {
  if (root === null) return null;
  let [1] = root;
  while ([2].[3] !== null) {
    [2] = [2].right;
  }
  return current.val;
}

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

Acurrent

Broot

Cright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'root' instead of a traversal variable.

Moving left instead of right.

We use a variable 'current' to traverse the tree moving right until no more right child exists.

Practice

1. In a Binary Search Tree (BST), where is the maximum element always located?

easy

A. At the rightmost node

B. At the leftmost node

C. At the root node

D. At any leaf node

Solution

1. Step 1: Understand BST property

   In a BST, left child nodes are smaller and right child nodes are larger than the parent node.

2. Step 2: Locate maximum element

   The maximum element is found by moving to the rightmost node because it has the largest value.

3. Final Answer:

   At the rightmost node -> Option A

4. Quick Check:

   Maximum element = rightmost node [OK]

Hint: Maximum is always the rightmost node in BST [OK]

Common Mistakes:

• Thinking maximum is at the root
• Confusing leftmost with maximum
• Assuming maximum can be any leaf

2. Which TypeScript code snippet correctly finds the maximum element in a BST node?

easy

A. while(node.right !== null) { node = node.right; } return node.value;

B. return Math.max(node.value, findMax(node.left), findMax(node.right));

C. if(node.left === null) return node.value; else return findMax(node.left);

D. while(node.left !== null) { node = node.left; } return node.value;

Solution

1. Step 1: Identify correct traversal direction

   To find maximum in BST, traverse right child nodes until no more right child exists.

2. Step 2: Check code correctness

   while(node.right !== null) { node = node.right; } return node.value; uses a while loop to move right until right child is null, then returns node value.

3. Final Answer:

   while(node.right !== null) { node = node.right; } return node.value; -> Option A

4. Quick Check:

   Traverse right until null, then return value [OK]

Hint: Traverse right child nodes until none left [OK]

Common Mistakes:

• Traversing left instead of right
• Using recursion incorrectly
• Using Math.max on subtree values (not BST property)

3. What will be the output of the following TypeScript code snippet?

class Node {
  constructor(public value: number, public left: Node | null = null, public right: Node | null = null) {}
}

function findMax(node: Node): number {
  while(node.right !== null) {
    node = node.right;
  }
  return node.value;
}

const root = new Node(10, new Node(5), new Node(15, null, new Node(20)));
console.log(findMax(root));

medium

A. 10

B. 5

C. 15

D. 20

Solution

1. Step 1: Understand tree structure

   Root is 10, left child 5, right child 15, and 15's right child is 20.

2. Step 2: Trace findMax function

   Function moves right from 10 to 15, then to 20, which has no right child, so returns 20.

3. Final Answer:

   20 -> Option D

4. Quick Check:

   Rightmost node value = 20 [OK]

Hint: Rightmost node value is maximum [OK]

Common Mistakes:

• Returning root value instead of rightmost
• Confusing left and right children
• Not following right child chain fully

4. The following TypeScript function is intended to find the maximum element in a BST. What is the error?

function findMax(node: Node): number {
  while(node.left !== null) {
    node = node.left;
  }
  return node.value;
}

medium

A. It returns undefined because node.value is missing

B. It causes infinite loop if node.left is null

C. It traverses left instead of right, so returns minimum instead of maximum

D. It throws error because Node type is incorrect

Solution

1. Step 1: Analyze traversal direction

   The function moves left child nodes, which leads to the minimum element in BST.

2. Step 2: Identify correct traversal for maximum

   Maximum element requires moving right child nodes, not left.

3. Final Answer:

   It traverses left instead of right, so returns minimum instead of maximum -> Option C

4. Quick Check:

   Left traversal = minimum, not maximum [OK]

Hint: Traverse right for max, left for min [OK]

Common Mistakes:

• Confusing left and right child traversal
• Assuming left traversal finds maximum
• Ignoring BST property of ordering

5. Given a BST where some nodes have only left children, how would you modify the findMax function to correctly find the maximum element even if the tree is skewed left?

hard

A. Return the root node's value directly since left skew means root is max

B. Traverse right child nodes until null; if none, return current node's value

C. Traverse left child nodes until null and return that value

D. Use recursion: return Math.max(node.value, findMax(node.left));

Solution

1. Step 1: Understand left skewed BST

   In a BST, values in left subtree < node <= values in right subtree. Even if some nodes have no right children, the maximum is found by traversing right until null.

2. Step 2: Modify findMax logic

   Traverse right child nodes until null; if no right child exists, current node is max.

3. Final Answer:

   Traverse right child nodes until null; if none, return current node's value -> Option B

4. Quick Check:

   Right traversal or current node if no right child [OK]

Hint: Check right child; if none, current node is max [OK]

Common Mistakes:

• Assuming root is always max in left skew
• Traversing left to find max
• Using recursion unnecessarily for simple right traversal