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DSA Typescriptprogramming~3 mins

Why BST Find Maximum Element in DSA Typescript?

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The Big Idea

What if you could find the biggest number instantly without checking every single one?

The Scenario

Imagine you have a big stack of books sorted by height, but they are all mixed up on a table. You want to find the tallest book quickly.

If you try to check each book one by one, it will take a long time and you might lose track.

The Problem

Checking every book manually is slow and tiring. You might miss some books or get confused about which is tallest.

Doing this with many numbers or data points by hand is even harder and error-prone.

The Solution

A Binary Search Tree (BST) organizes data so that the biggest item is always easy to find by just following one path.

Finding the maximum means going right until you can't go further, making it fast and simple.

Before vs After
✗ Before
function findMaxManual(arr: number[]): number {
  let max = arr[0];
  for (let i = 1; i < arr.length; i++) {
    if (arr[i] > max) max = arr[i];
  }
  return max;
}
✓ After
function findMaxBST(root: TreeNode | null): number | null {
  if (!root) return null;
  let current = root;
  while (current.right) {
    current = current.right;
  }
  return current.value;
}
What It Enables

This lets you quickly find the largest value in a sorted structure without checking every item.

Real Life Example

In a leaderboard for a game, finding the highest score fast is important. A BST helps find the top score quickly.

Key Takeaways

Manual search is slow and error-prone.

BST structure makes max value easy to find by going right.

Efficient for large sorted data sets.

Practice

(1/5)
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
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