BST Find Maximum Element in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the maximum value in a BST.

DSA Python

def find_max(root):
    current = root
    while current.[1] is not None:
        current = current.right
    return current.data

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

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

Trying to access a non-existent attribute.

Not checking if the root is None before traversal.

The maximum element in a BST is found by going to the rightmost node. So, we keep moving to the right child until there is no right child left.

2fill in blank

medium

Complete the code to handle the case when the BST is empty.

DSA Python

def find_max(root):
    if [1] is None:
        return None
    current = root
    while current.right is not None:
        current = current.right
    return current.data

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

Aroot.right

Bcurrent

Croot

Droot.left

Attempts:

3 left

💡 Hint

Common Mistakes

Checking current instead of root for None.

Not returning anything when the tree is empty.

Trying to access attributes of None.

We check if the root itself is None to handle an empty tree. If it is None, we return None as there is no maximum element.

3fill in blank

hard

Fix the error in the loop condition to correctly find the maximum element.

DSA Python

def find_max(root):
    current = root
    while current.[1] != None:
        current = current.right
    return current.data

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

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' in the loop condition.

Using '!=' None instead of 'is not None' (though both work, 'is not None' is preferred).

Not updating current inside the loop.

The loop should continue while the right child exists. Using 'right' ensures we move to the rightmost node which holds the maximum value.

4fill in blank

hard

Fill both blanks to create a function that returns the maximum value or None if the tree is empty.

DSA Python

def find_max(root):
    if root is [1]:
        return [2]
    current = root
    while current.right is not None:
        current = current.right
    return current.data

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

ANone

Broot

CFalse

DTrue

Attempts:

3 left

💡 Hint

Common Mistakes

Returning root instead of None when tree is empty.

Checking root == False instead of root is None.

Not handling empty tree case.

If the root is None, it means the tree is empty, so we return None. This handles the empty tree case safely.

5fill in blank

hard

Fill all three blanks to create a safe function that finds the maximum element in a BST.

DSA Python

def find_max(root):
    if [1] is None:
        return [2]
    current = [3]
    while current.right is not None:
        current = current.right
    return current.data

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

Aroot

BNone

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Starting traversal from current instead of root.

Not returning None when tree is empty.

Using wrong variable names.

First, check if root is None to handle empty tree. Return None if empty. Then start traversal from root to find the rightmost node which holds the maximum value.

Practice

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

easy

A. At the root node

B. At the leftmost node

C. At the rightmost node

D. At any leaf node

Solution

1. Step 1: Understand BST property

   In a BST, all values in the right subtree are greater than the root, so the largest value is towards the right.

2. Step 2: Locate the maximum element

   The maximum element is found by moving to the right child repeatedly until no more right child exists.

3. Final Answer:

   At the rightmost node -> Option C

4. Quick Check:

   Maximum element = rightmost node [OK]

Hint: Go right until no more right child [OK]

Common Mistakes:

• Thinking maximum is at the root
• Looking at left subtree for maximum
• Assuming maximum can be any leaf

2. Which of the following Python code snippets correctly finds the maximum element in a BST node structure where each node has left and right attributes?

easy

A. def find_max(root):
while root.right:
root = root.right
return root.data

B. def find_max(root):
while root.left:
root = root.left
return root.data

C. def find_max(root):
if root is None:
return None
return root.data

D. def find_max(root):
return max(root.left.data, root.right.data)

Solution

1. Step 1: Check traversal direction for max

   Maximum is found by moving right until no right child exists, so loop should check root.right.

2. Step 2: Verify code correctness

   def find_max(root): while root.right: root = root.right return root.data correctly loops while root.right exists, then returns root.data. Others either go left or incorrectly use max function.

3. Final Answer:

   while root.right: root = root.right; return root.data -> Option A

4. Quick Check:

   Loop right child until none = correct max finder [OK]

Hint: Loop on root.right to find max [OK]

Common Mistakes:

• Looping on root.left instead of root.right
• Returning root.data without traversal
• Using max() on children without recursion

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

class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

def find_max(root):
    current = root
    while current.right:
        current = current.right
    return current.data

root = Node(10)
root.left = Node(5)
root.right = Node(20)
root.right.right = Node(30)
print(find_max(root))

medium

A. 10

B. 30

C. 20

D. 5

Solution

1. Step 1: Trace the BST structure

   Root is 10, right child 20, right of 20 is 30. Maximum is the rightmost node 30.

2. Step 2: Follow the find_max function

   Function moves right from 10 to 20, then to 30, then stops as 30 has no right child, returning 30.

3. Final Answer:

   30 -> Option B

4. Quick Check:

   Rightmost node data = 30 [OK]

Hint: Follow right pointers to last node [OK]

Common Mistakes:

• Returning root data instead of rightmost
• Confusing left and right children
• Stopping traversal too early

4. The following code is intended to find the maximum element in a BST but contains a bug. What is the bug?

def find_max(root):
    if root is None:
        return None
    while root.left:
        root = root.left
    return root.data

medium

A. It causes infinite loop

B. It does not handle empty tree

C. It returns None instead of data

D. It incorrectly traverses left child instead of right child

Solution

1. Step 1: Analyze traversal direction

   The code loops while root.left exists, moving left, but maximum is on the right side in BST.

2. Step 2: Identify correct traversal

   To find maximum, loop should be on root.right until no right child remains.

3. Final Answer:

   It incorrectly traverses left child instead of right child -> Option D

4. Quick Check:

   Max element found by going right, not left [OK]

Hint: Max is rightmost, so loop on root.right [OK]

Common Mistakes:

• Looping on left child for max
• Ignoring empty tree check
• Returning wrong node data

5. Given a BST where some nodes have duplicate values, which approach correctly finds the maximum element considering duplicates are allowed only on the right subtree?

class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

def find_max(root):
    # Your code here
    pass

Choose the correct implementation to find the maximum element.

hard

A. def find_max(root):
current = root
while current.right:
current = current.right
return current.data

B. def find_max(root):
if root.right is None:
return root.data
return find_max(root.left)

C. def find_max(root):
max_val = root.data
max_val = max(max_val, find_max(root.left))
max_val = max(max_val, find_max(root.right))
return max_val

D. def find_max(root):
if root is None:
return float('-inf')
return max(root.data, find_max(root.left))

Solution

1. Step 1: Understand duplicates in BST

   Duplicates allowed only on right means maximum is still the rightmost node.

2. Step 2: Choose correct traversal

   def find_max(root): current = root while current.right: current = current.right return current.data correctly moves right until no right child exists, returning the maximum value.

3. Step 3: Analyze other options

   def find_max(root): if root.right is None: return root.data return find_max(root.left) incorrectly recurses left; def find_max(root): max_val = root.data max_val = max(max_val, find_max(root.left)) max_val = max(max_val, find_max(root.right)) return max_val lacks None checks before recursing, causing AttributeError on None children; def find_max(root): if root is None: return float('-inf') return max(root.data, find_max(root.left)) ignores right subtree.

4. Final Answer:

   while current.right: current = current.right; return current.data -> Option A

5. Quick Check:

   Rightmost node = max even with duplicates [OK]

Hint: Duplicates on right? Still go rightmost [OK]

Common Mistakes:

• Searching left subtree for max
• Checking both subtrees unnecessarily
• Ignoring None checks causing errors