BST Find Minimum Element in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A Binary Search Tree is a tree data structure where each node has at most two children. For every node, all nodes in its left subtree have smaller values, and all nodes in its right subtree have larger values.

beginner

How do you find the minimum element in a BST?

To find the minimum element in a BST, start at the root and keep moving to the left child until you reach a node with no left child. That node contains the minimum value.

beginner

Why does moving left in a BST lead to the minimum element?

Because in a BST, left children always have smaller values than their parent nodes. So, moving left repeatedly leads to the smallest value.

intermediate

What is the time complexity of finding the minimum element in a BST?

The time complexity is O(h), where h is the height of the BST. In the best case (balanced tree), it's O(log n). In the worst case (skewed tree), it's O(n).

beginner

Show the Python code snippet to find the minimum element in a BST node.

def find_min(node): current = node while current.left is not None: current = current.left return current.data

In a BST, where is the minimum element located?

A At the leftmost node

B At the rightmost node

C At the root node

D At any leaf node

The minimum element is always at the leftmost node because left children have smaller values.

What should you do if the BST is empty when finding the minimum element?

A Return the rightmost node

B Return the root node

C Return None or indicate no minimum

D Throw an error always

If the BST is empty, there is no minimum element, so returning None or a similar indication is correct.

What is the first step to find the minimum element in a BST?

A Start at the middle node

B Start at the right child

C Start at the leaf node

D Start at the root node

You always start at the root and move left to find the minimum.

If a BST has only one node, what is the minimum element?

A The single node itself

B No minimum element

C The left child

D The right child

With only one node, that node is both the minimum and maximum.

What is the time complexity to find the minimum element in a balanced BST with n nodes?

A O(n)

B O(log n)

C O(1)

D O(n log n)

In a balanced BST, height h is about log n, so finding minimum takes O(log n).

Explain step-by-step how to find the minimum element in a BST.

Think about the BST property and left children.

Write a simple Python function to find the minimum element in a BST and explain how it works.

Use a while loop and check left child.

Practice

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

easy

A. At the leftmost node

B. At the rightmost node

C. At the root node

D. At any leaf node

Solution

1. Step 1: Understand BST property

   In a BST, left child nodes have smaller values than their parents, so the smallest value is always towards the left.

2. Step 2: Locate minimum element

   By moving left from the root repeatedly until no more left child exists, we reach the minimum element.

3. Final Answer:

   At the leftmost node -> Option A

4. Quick Check:

   Minimum element = leftmost node [OK]

Hint: Minimum is always the leftmost node in BST [OK]

Common Mistakes:

• Confusing minimum with root node
• Thinking minimum is at rightmost node
• Assuming minimum can be any leaf

2. Which of the following Python code snippets correctly finds the minimum element in a BST node named root?

easy

A. while root.right is not None:
root = root.right
print(root.val)

B. while root.left is not None:
root = root.left
print(root.val)

C. if root.left is None:
print(root.val)

D. print(min(root.val, root.left.val))

Solution

1. Step 1: Check loop for minimum search

   while root.left is not None: root = root.left print(root.val) correctly moves left while left child exists, which is how to find minimum in BST.

2. Step 2: Verify other options

   while root.right is not None: root = root.right print(root.val) moves right (wrong for minimum), C only checks once, D uses min on values without traversal.

3. Final Answer:

   while root.left is not None: root = root.left print(root.val) -> Option B

4. Quick Check:

   Traverse left until None = while root.left is not None: root = root.left print(root.val) [OK]

Hint: Traverse left child until None to find minimum [OK]

Common Mistakes:

• Traversing right child instead of left
• Not using a loop to reach leftmost node
• Using min() function incorrectly on nodes

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

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

def find_min(root):
    while root.left:
        root = root.left
    return root.val

root = Node(10)
root.left = Node(5)
root.right = Node(15)
root.left.left = Node(2)
root.left.right = Node(7)

print(find_min(root))

medium

A. 5

B. 10

C. 7

D. 2

Solution

1. Step 1: Trace the BST structure

   The tree root is 10, left child 5, left-left child 2, left-right child 7, right child 15.

2. Step 2: Follow find_min function

   The function moves left from 10 to 5, then left again to 2, which has no left child, so returns 2.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Leftmost node value = 2 [OK]

Hint: Minimum is leftmost node value, here 2 [OK]

Common Mistakes:

• Returning root value instead of leftmost
• Confusing left-right child as minimum
• Not following loop correctly

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

def find_min(root):
    while root.right:
        root = root.right
    return root.val

medium

A. It traverses right child instead of left child

B. It returns root.val without checking children

C. It causes infinite loop if root.right is None

D. It does not handle empty tree (None root)

Solution

1. Step 1: Analyze traversal direction

   The code moves to the right child repeatedly, which finds the maximum, not minimum.

2. Step 2: Identify correct traversal for minimum

   Minimum element is found by moving left child repeatedly, so the code should use root.left.

3. Final Answer:

   It traverses right child instead of left child -> Option A

4. Quick Check:

   Minimum = traverse left, not right [OK]

Hint: Minimum is leftmost, so traverse left child [OK]

Common Mistakes:

• Using right child traversal for minimum
• Ignoring empty tree case
• Assuming root is minimum without traversal

5. Given a BST where some nodes may have duplicate values on the left subtree, which approach correctly finds the minimum element?

def find_min(root):
    # Your code here

Choose the correct code snippet to handle duplicates on the left side.

hard

A. while root.right:
root = root.right
return root.val

B. while root.left and root.left.val < root.val:
root = root.left
return root.val

C. while root.left:
root = root.left
return root.val

D. if root.left and root.left.val < root.val:
return find_min(root.left)
return root.val

Solution

1. Step 1: Understand duplicates on left subtree

   In BSTs allowing duplicates on left (e.g., left <= parent), the minimum is always the leftmost node.

2. Step 2: Correct approach

   while root.left: root = root.left return root.val unconditionally traverses left to the leftmost node, handling duplicates correctly.

3. Step 3: Why others fail

   A finds maximum. C and D only traverse if strictly <, stopping early on equal values, missing smaller nodes further left.

4. Final Answer:

   while root.left: root = root.left return root.val -> Option C

5. Quick Check:

   Unconditional left traversal finds min even with duplicates [OK]

Hint: Always traverse left until None, even with duplicates [OK]

Common Mistakes:

• Ignoring duplicates and returning first left child
• Moving right instead of left
• Stopping traversal too early