BST Property and Why It Matters in DSA Javascript - Build from Scratch

BST Property and Why It Matters

📖 Scenario: Imagine you are organizing a collection of books by their unique ID numbers. You want to store them so that searching for a book by its ID is very fast. A Binary Search Tree (BST) helps you do this by keeping books in a special order.

🎯 Goal: You will build a simple Binary Search Tree (BST) with three nodes and then print the tree structure to see how the BST property keeps the data organized.

📋 What You'll Learn

Create three nodes with values 10, 5, and 15

Link the nodes to form a BST where 5 is left child of 10 and 15 is right child of 10

Print the tree nodes in order: left child, root, right child

💡 Why This Matters

🌍 Real World

BSTs are used in databases and file systems to quickly find data by keeping it sorted.

💼 Career

Understanding BSTs helps in software development roles that involve data storage, search optimization, and algorithm design.

1

Create BST nodes

Create three variables called root, leftChild, and rightChild with values 10, 5, and 15 respectively.

DSA Javascript

# Your code here

Hint

Use const to create variables and assign the exact numbers.

2

Link nodes to form BST

Create an object called bst with properties value, left, and right. Set value to root, left to leftChild, and right to rightChild.

DSA Javascript

const root = 10;
const leftChild = 5;
const rightChild = 15;

# Your code here

Hint

Create an object with keys value, left, and right using the variables from Step 1.

3

Write function to print BST in order

Write a function called inOrderPrint that takes a node object and prints its left child, then its value, then its right child. Use console.log to print values.

DSA Javascript

const root = 10;
const leftChild = 5;
const rightChild = 15;

const bst = {
  value: root,
  left: leftChild,
  right: rightChild
};

# Your code here

Hint

Print left child first, then value, then right child using console.log.

4

Print the BST nodes in order

Call the function inOrderPrint with the bst object to print the nodes in order.

DSA Javascript

const root = 10;
const leftChild = 5;
const rightChild = 15;

const bst = {
  value: root,
  left: leftChild,
  right: rightChild
};

function inOrderPrint(node) {
  if (node.left !== null) {
    console.log(node.left);
  }
  console.log(node.value);
  if (node.right !== null) {
    console.log(node.right);
  }
}

# Your code here

Hint

Call inOrderPrint(bst); to see the nodes printed in order.

Practice

1. What is the main property that defines a Binary Search Tree (BST)?

easy

A. Each node has exactly two children.

B. For each node, all values in the left subtree are smaller, and all values in the right subtree are larger.

C. All nodes have the same value.

D. Nodes are arranged in a circular linked list.

Solution

1. Step 1: Understand BST node value rules

   In a BST, the left subtree contains values smaller than the node, and the right subtree contains values larger than the node.

2. Step 2: Compare options with BST definition

   For each node, all values in the left subtree are smaller, and all values in the right subtree are larger. matches the BST property exactly, while others describe different structures.

3. Final Answer:

   For each node, all values in the left subtree are smaller, and all values in the right subtree are larger. -> Option B

4. Quick Check:

   BST property = left < node < right [OK]

Hint: Remember: left < node < right defines BST [OK]

Common Mistakes:

• Confusing BST with binary tree (not ordered)
• Thinking all nodes must have two children
• Assuming all values are equal

2. Which of the following is the correct way to check if a value should go to the left subtree in a BST insertion function in JavaScript?

easy

A. if (value < node.value) { insertLeft(); }

B. if (value > node.value) { insertLeft(); }

C. if (value == node.value) { insertLeft(); }

D. if (value >= node.value) { insertLeft(); }

Solution

1. Step 1: Recall BST insertion rule

   Values smaller than the current node go to the left subtree.

2. Step 2: Match condition to insertion side

   Only the condition 'value < node.value' correctly directs insertion to the left.

3. Final Answer:

   if (value < node.value) { insertLeft(); } -> Option A

4. Quick Check:

   Smaller values go left [OK]

Hint: Use < for left subtree insertion [OK]

Common Mistakes:

• Using > instead of < for left subtree
• Using == which is not for left or right
• Using >= which includes equal values incorrectly

3. What will be the in-order traversal output of the following BST?

    10
   /  \
  5   15
 / \    \
3   7    20

medium

A. [20, 15, 10, 7, 5, 3]

B. [10, 5, 3, 7, 15, 20]

C. [3, 5, 7, 10, 15, 20]

D. [5, 3, 7, 10, 15, 20]

Solution

1. Step 1: Understand in-order traversal

   In-order traversal visits left subtree, then node, then right subtree recursively.

2. Step 2: Apply in-order to given tree

   Left subtree of 10: 3, 5, 7; then 10; then right subtree: 15, 20. Combined: [3, 5, 7, 10, 15, 20]

3. Final Answer:

   [3, 5, 7, 10, 15, 20] -> Option C

4. Quick Check:

   In-order = sorted BST values [OK]

Hint: In-order traversal of BST gives sorted values [OK]

Common Mistakes:

• Confusing pre-order or post-order with in-order
• Listing nodes in insertion order
• Reversing the traversal order

4. Consider this BST insertion code snippet in JavaScript:

function insert(node, value) {
  if (!node) return { value, left: null, right: null };
  if (value <= node.value) {
    node.left = insert(node.left, value);
  } else {
    node.right = insert(node.right, value);
  }
  return node;
}

What is the main issue with this code regarding BST property?

medium

A. It inserts larger values on the left subtree incorrectly.

B. It does not handle the case when node is null.

C. It returns undefined instead of the node.

D. It allows duplicate values on the left subtree, which may break strict BST rules.

Solution

1. Step 1: Analyze duplicate handling in insertion

   The condition 'value <= node.value' inserts duplicates to the left subtree.

2. Step 2: Understand BST strictness

   Strict BSTs require left subtree values strictly less than node value; duplicates on left can break this.

3. Final Answer:

   It allows duplicate values on the left subtree, which may break strict BST rules. -> Option D

4. Quick Check:

   Duplicates on left break strict BST [OK]

Hint: Use < for left to avoid duplicates on left [OK]

Common Mistakes:

• Ignoring duplicate value placement
• Thinking null check is missing
• Confusing left and right subtree insertion

5. You have a BST where some nodes violate the BST property because their left child has a value greater than the node. Which approach best fixes the tree without rebuilding it completely?

hard

A. Swap the values of the violating nodes with their children to restore BST property locally.

B. Ignore the violations as they do not affect search operations.

C. Delete all nodes and insert them again in correct order.

D. Perform an in-order traversal, collect values, then rebuild the BST from sorted values.

Solution

1. Step 1: Understand the violation type

   Left child having greater value than node breaks BST property locally.

2. Step 2: Choose minimal fix without full rebuild

   Swapping values of violating nodes with their children can restore local BST property without full rebuild.

3. Step 3: Evaluate other options

   Perform an in-order traversal, collect values, then rebuild the BST from sorted values. rebuilds fully, Delete all nodes and insert them again in correct order. deletes all nodes (costly), Ignore the violations as they do not affect search operations. ignores problem (wrong).

4. Final Answer:

   Swap the values of the violating nodes with their children to restore BST property locally. -> Option A

5. Quick Check:

   Local swaps fix BST violations efficiently [OK]

Hint: Fix BST by swapping violating node values locally [OK]

Common Mistakes:

• Rebuilding entire tree unnecessarily
• Ignoring violations thinking no impact
• Deleting all nodes instead of fixing