Searching in BST in Data Structures Theory - Mini Project: Build & Apply

Searching in BST

📖 Scenario: You have a collection of numbers stored in a special tree called a Binary Search Tree (BST). This tree helps you find numbers quickly by following simple rules.Imagine you want to check if a certain number is in your collection.

🎯 Goal: Build a step-by-step process to search for a number in a Binary Search Tree (BST) by following the BST rules.

📋 What You'll Learn

Create a simple BST structure with nodes and values

Set a target value to search for

Use the BST search logic to find the target

Complete the search by returning the result

💡 Why This Matters

🌍 Real World

BSTs are used in databases and file systems to quickly find data without searching everything.

💼 Career

Understanding BST search is important for software developers working with data structures and algorithms.

1

Create the BST nodes

Create a dictionary called bst representing a simple BST with this structure: root node value 8, left child value 3, right child value 10. Use nested dictionaries with keys value, left, and right. Set children to None if no child exists.

Data Structures Theory

# Create the bst dictionary with root 8, left child 3, right child 10
# Your code here

Hint

Use nested dictionaries to represent nodes. Each node has 'value', 'left', and 'right' keys.

2

Set the target value to search

Create a variable called target and set it to the number 10 which you want to find in the BST.

Data Structures Theory

bst = {
    'value': 8,
    'left': {'value': 3, 'left': None, 'right': None},
    'right': {'value': 10, 'left': None, 'right': None}
}
# Set the target value to 10
# Your code here

Hint

Just assign the number 10 to the variable named target.

3

Implement the BST search logic

Write a function called search_bst that takes two arguments: node and target. Use the BST search rules: if node is None, return False. If node['value'] equals target, return True. If target is less than node['value'], search the left child. Otherwise, search the right child.

Data Structures Theory

bst = {
    'value': 8,
    'left': {'value': 3, 'left': None, 'right': None},
    'right': {'value': 10, 'left': None, 'right': None}
}
target = 10

# Define the search_bst function using BST rules
# Your code here

Hint

Use recursion to check left or right child based on comparison.

4

Complete the search by calling the function

Create a variable called found and set it to the result of calling search_bst with bst and target as arguments.

Data Structures Theory

bst = {
    'value': 8,
    'left': {'value': 3, 'left': None, 'right': None},
    'right': {'value': 10, 'left': None, 'right': None}
}
target = 10

def search_bst(node, target):
    if node is None:
        return False
    if node['value'] == target:
        return True
    if target < node['value']:
        return search_bst(node['left'], target)
    else:
        return search_bst(node['right'], target)

# Store the search result in found
# Your code here

Hint

Call the function with the root node and target, save the result in found.

Practice

1. What is the main advantage of searching in a Binary Search Tree (BST)?

easy

A. It allows faster search by using the order of elements

B. It stores elements in random order for quick access

C. It uses hashing to find elements instantly

D. It searches all nodes one by one sequentially

Solution

1. Step 1: Understand BST property

   A BST keeps elements ordered so smaller values are on the left and larger on the right.

2. Step 2: Use order for searching

   This order lets us decide to go left or right, skipping half the tree each step, making search faster.

3. Final Answer:

   It allows faster search by using the order of elements -> Option A

4. Quick Check:

   BST order speeds search = It allows faster search by using the order of elements [OK]

Hint: BST order guides search direction quickly [OK]

Common Mistakes:

• Thinking BST stores elements randomly
• Confusing BST with hashing
• Assuming linear search in BST

2. Which of the following is the correct way to decide the next node to visit when searching for a value in a BST?

easy

A. Go left if target is greater than current node

B. Go left if target is smaller than current node

C. Go right if target is smaller than current node

D. Always go to the root node

Solution

1. Step 1: Recall BST search rule

   If the target is smaller than the current node's value, we move to the left child.

2. Step 2: Apply rule to options

   Go left if target is smaller than current node correctly states to go left if target is smaller, which matches BST property.

3. Final Answer:

   Go left if target is smaller than current node -> Option B

4. Quick Check:

   Smaller target -> left child = Go left if target is smaller than current node [OK]

Hint: Smaller target means go left in BST [OK]

Common Mistakes:

• Reversing left and right directions
• Ignoring BST ordering rules
• Always going to root node

3. Consider the BST below:

      15
     /  \
    10  20
   /    / \
  8    17 25

Which nodes will be visited when searching for the value 17?

medium

A. [10, 8, 17]

B. [15, 10, 8]

C. [15, 20, 25]

D. [15, 20, 17]

Solution

1. Step 1: Start at root and compare with 17

   Root is 15. Since 17 > 15, move right to 20.

2. Step 2: Compare 20 with 17

   17 < 20, so move left to 17, which matches the target.

3. Final Answer:

   [15, 20, 17] -> Option D

4. Quick Check:

   Path to 17 = 15 -> 20 -> 17 [OK]

Hint: Follow BST rules: left if smaller, right if larger [OK]

Common Mistakes:

• Going left from 15 when target is larger
• Skipping nodes in path
• Confusing node values

4. You wrote code to search a BST but it always returns None even when the value exists. What is the most likely mistake?

medium

A. Not moving to left child when target is smaller

B. Using a queue instead of recursion

C. Always moving to left child regardless of target

D. Checking only the root node

Solution

1. Step 1: Understand BST search logic

   Search must move left if target is smaller, right if larger.

2. Step 2: Identify error in always moving left

   If code always moves left, it misses nodes on the right side where target might be.

3. Final Answer:

   Always moving to left child regardless of target -> Option C

4. Quick Check:

   Wrong direction causes search failure = Always moving to left child regardless of target [OK]

Hint: Move left or right based on comparison, not always left [OK]

Common Mistakes:

• Ignoring right subtree
• Checking only root node
• Using wrong data structure for traversal

5. Given a BST where some nodes have duplicate values on the right subtree, how should the search algorithm be adapted to find all occurrences of a target value?

hard

A. Traverse both left and right subtrees when node equals target

B. Search left subtree only once target is found

C. Stop search immediately after first match

D. Ignore duplicates and return first found

Solution

1. Step 1: Understand duplicates in BST

   Duplicates are stored in right subtree, so multiple matches can exist there.

2. Step 2: Adapt search to find all matches

   When a node equals target, search both left (for smaller) and right (for duplicates) subtrees to find all occurrences.

3. Final Answer:

   Traverse both left and right subtrees when node equals target -> Option A

4. Quick Check:

   Check both sides for duplicates = Traverse both left and right subtrees when node equals target [OK]

Hint: Check both subtrees when value matches to find duplicates [OK]

Common Mistakes:

• Stopping after first match
• Ignoring right subtree duplicates
• Searching only one subtree