BST Find Maximum Element in DSA Python - Execution Trace

Concept Flow - BST Find Maximum Element

Start at root node

↓

Check if right child exists?

No→Current node is max

Yes↓

Move to right child

↩Back to check right child

Start at the root and keep moving right until no right child exists; that node holds the maximum value.

Execution Sample

DSA Python

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

This code finds the maximum value in a BST by moving right until no right child is found.

Execution Table

Step	Operation	Current Node	Right Child Exists?	Pointer Changes	Visual State
1	Start at root	50	Yes	current = 50	50 / \ 30 70 / \ 60 80
2	Check right child	50	Yes	Move current to 70	50 / \ 30 70 / \ 60 80
3	Check right child	70	Yes	Move current to 80	50 / \ 30 70 / \ 60 80
4	Check right child	80	No	Stop, current is max	50 / \ 30 70 / \ 60 80
5	Return value	80	N/A	Return 80	80 is max element

💡 No right child at node 80, so 80 is the maximum element.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
current	root (50)	50	70	80	80

Key Moments - 3 Insights

Why do we keep moving to the right child to find the maximum?

In a BST, all right children have values greater than their parent. The execution_table rows 2 and 3 show moving right to nodes 70 and 80, increasing the value.

What if the root has no right child?

If the root has no right child, the root itself is the maximum. The execution_table row 4 shows stopping when no right child exists.

Why do we stop when current.right is None, not when current is None?

Stopping at current.right == None means current is the rightmost node. The code and execution_table row 4 confirm this stopping condition.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'current' after step 2?

A 50

B 80

C 70

D 30

💡 Hint

Check the 'Pointer Changes' and 'Current Node' columns in execution_table row 2.

At which step does the algorithm stop moving right because no right child exists?

A Step 3

B Step 4

C Step 2

D Step 1

💡 Hint

Look at the 'Right Child Exists?' column in execution_table to find when it becomes 'No'.

If the node 70 had no right child, what would be the maximum element?

A 70

B 80

C 60

D 50

💡 Hint

Refer to the concept_flow and execution_table steps where the right child is checked.

Concept Snapshot

BST Find Maximum Element:
- Start at root node
- Move right while right child exists
- When no right child, current node is max
- Return current node's value
- Right subtree holds larger values in BST

Full Transcript

To find the maximum element in a Binary Search Tree (BST), start at the root node. Check if the current node has a right child. If yes, move to that right child and repeat. Continue moving right until you reach a node with no right child. That node contains the maximum value in the BST. This works because in BSTs, right children always have greater values than their parents. The code uses a loop to move right until no right child is found, then returns the current node's value as the maximum.

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