Kth Smallest Element in BST | Optimized Using BST Iterator logic | Python, Java & C++ | DSA

In this video, we solve the Kth Smallest Element in a Binary Search Tree (BST) problem using an optimized approach based on the BST Iterator concept. Since the inorder traversal of a BST gives elements in sorted order, the Kth element in inorder traversal is the Kth smallest element. What you'll learn: How to find the Kth smallest element in a BST Why BST inorder traversal gives sorted order Basic recursive approach Optimized approach using BST Iterator How to simulate inorder traversal using a stack Why we don't need to traverse the entire tree Step-by-step dry run Python, Java & C++ implementation Time and Space Complexity Important BST interview concepts Key Idea Instead of storing the complete inorder traversal, we build a BST Iterator that returns the next smallest element whenever required. The iterator uses a stack to simulate: Left → Root → Right We repeatedly call next() and count the elements. When the count reaches K, we have found the Kth smallest element. Optimized Approach Time: O(H + K) Space: O(H) where H is the height of the BST. The BST Iterator makes the solution more efficient when we only need the first K elements instead of generating the complete sorted traversal. This problem is an important combination of: BST + Inorder Traversal + Stack + BST Iterator and is a very useful pattern for coding interviews. Rising Brain (DSA Sheet): https://www.risingbrain.org/sheet Connect with me here: LinkedIn –   / anjalikumari22   Instagram –   / rbanjali.codes   Twitter (X) – https://x.com/anjali1kumari?s=21