BST Find Maximum Element in DSA Python

Mental Model

In a Binary Search Tree, the biggest number is always found by going as far right as possible.

Analogy: Imagine a family tree where older generations are on the left and younger generations are on the right. The youngest person is the one farthest to the right.

      10
     /  \
    5    15
         / \
        12  20
               ↑ (max is here)

Dry Run Walkthrough

Input: BST: 10 -> 5, 15 -> 12, 20; find max element

Goal: Find the largest value in the BST by moving right until no more right child exists

Step 1: Start at root node 10

10 -> [curr->15] -> 20
 ↑ (curr at 10)

Why: We start at the root to find the maximum

Step 2: Move curr pointer to right child 15

10 -> 15 -> [curr->20]
       ↑ (curr at 15)

Why: Maximum is always in the right subtree, so move right

Step 3: Move curr pointer to right child 20

10 -> 15 -> 20
             ↑ (curr at 20)

Why: Keep moving right until no right child

Step 4: No right child of 20, stop and return 20

10 -> 15 -> 20
             ↑ (curr at 20)

Why: 20 is the maximum value in the BST

Result:

10 -> 15 -> 20
             ↑ (max = 20)

Annotated Code

DSA Python

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

def find_max(root):
    if root is None:
        return None
    curr = root
    while curr.right is not None:
        curr = curr.right  # move right to find max
    return curr.val

# Driver code
root = Node(10)
root.left = Node(5)
root.right = Node(15)
root.right.left = Node(12)
root.right.right = Node(20)

max_value = find_max(root)
print(max_value)

while curr.right is not None:

advance curr to right child to find larger values

curr = curr.right # move right to find max

move curr pointer right until no more right child

return curr.val

return value of the rightmost node as maximum

Output

20

Complexity Analysis

Time: O(h) because we only move down the right children where h is tree height

Space: O(1) because we use only a pointer variable without extra storage

vs Alternative: Compared to traversing all nodes (O(n)), this is faster as it uses BST property to go directly right

Edge Cases

Empty tree (root is None)

Returns None since there is no maximum

DSA Python

if root is None:
        return None

Tree with only one node

Returns the single node's value as maximum

DSA Python

while curr.right is not None:

Tree where all nodes are left children (no right child)

Returns root value as maximum since no right child exists

DSA Python

while curr.right is not None:

When to Use This Pattern

When asked to find the largest value in a BST, look for the rightmost node because BSTs store larger values to the right.

Common Mistakes

Mistake: Trying to check both left and right children instead of only moving right
Fix: Only move right child pointers to find maximum in BST

Mistake: Not handling empty tree and causing error
Fix: Add a check for root being None before traversal

Summary

Finds the largest value in a BST by moving right until no right child exists.

Use when you need the maximum element quickly in a BST.

The maximum is always the rightmost node in a BST.

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