What is the time complexity of Kth Smallest Element in a BST?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Kth Smallest Element in a BST?

This problem uses Tree, Depth-First Search, Binary Search Tree, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Kth Smallest Element in a BST?

Insight 1: An in-order traversal of a BST yields the nodes in ascending order (sorted order). This is the fundamental property we leverage. Insight 2: Iterative in-order traversal using a stack avoids recursion, potentially saving space in certain scenarios.

Which companies ask LeetCode 230. Kth Smallest Element in a BST in interviews?

This problem is frequently asked at Agoda, Cisco, Google, Oracle.

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Kth Smallest Element in a BST - LeetCode 230 Solution

Q: How to solve Kth Smallest Element in a BST?

The solution employs an iterative in-order traversal of the BST using a stack. We maintain a stack to simulate the recursive calls of a standard in-order traversal. We traverse as far left as possible from the current node, pushing each node onto the stack. Then, we pop a node from the stack (which represents the next smallest element) and decrement `k`. If `k` becomes 0, we've found the k-th smallest element, and we return its value. Otherwise, we move to the right child of the popped node and repeat the process.

Kth Smallest Element in a BST - Complete Solution Guide

Kth Smallest Element in a BST is LeetCode problem 230, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given the root of a binary search tree, and an integer k , return the k th smallest value ( 1-indexed ) of all the values of the nodes in the tree . Example 1: Input: root = [3,1,4,null,2], k = 1 Output: 1 Example 2: Input: root = [5,3,6,2,4,null,null,1], k = 3 Output: 3 Constraints: The number of nodes in the tree is n . 1 <= k <= n <= 10 4 0 <= Node.val <= 10 4 Follow up: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequentl

Detailed Explanation

The problem asks us to find the k-th smallest element in a Binary Search Tree (BST). The values are 1-indexed, meaning the smallest element is the 1st smallest, the second smallest is the 2nd smallest, and so on. A BST has the property that for any node, all nodes in its left subtree have values smaller than the node's value, and all nodes in its right subtree have values greater than the node's value. The input is the root of the BST and an integer `k`. The output is the value of the k-th smallest node in the BST.

Solution Approach

The solution employs an iterative in-order traversal of the BST using a stack. We maintain a stack to simulate the recursive calls of a standard in-order traversal. We traverse as far left as possible from the current node, pushing each node onto the stack. Then, we pop a node from the stack (which represents the next smallest element) and decrement `k`. If `k` becomes 0, we've found the k-th smallest element, and we return its value. Otherwise, we move to the right child of the popped node and repeat the process.

Step-by-Step Algorithm

Step 1: Initialize an empty stack and set the current node to the root of the BST.
Step 2: While the current node is not null, push the current node onto the stack and move to its left child. This process continues until the leftmost node is reached.
Step 3: If the stack is empty, the traversal is complete. Otherwise, pop a node from the stack.
Step 4: Decrement `k`. If `k` is 0, return the value of the popped node (the k-th smallest element).
Step 5: Set the current node to the right child of the popped node, and repeat steps 2-4.

Key Insights

Insight 1: An in-order traversal of a BST yields the nodes in ascending order (sorted order). This is the fundamental property we leverage.
Insight 2: Iterative in-order traversal using a stack avoids recursion, potentially saving space in certain scenarios.
Insight 3: We can stop the in-order traversal once we have found the k-th smallest element, avoiding processing the entire tree.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Tree, Depth-First Search, Binary Search Tree, Binary Tree.

Companies

Asked at: Agoda, Cisco, Google, Oracle.

Kth Smallest Element in a BST | LeetCode 230

Medium

Tree Depth-First Search Binary Search Tree Binary Tree

Time: O(n)Space: O(n)

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        java.util.Stack<TreeNode> stack = new java.util.Stack<>();
        TreeNode current = root;
        
        while (current != null || !stack.isEmpty()) {
            while (current != null) {
                stack.push(current);
                current = current.left;
            }
            
            current = stack.pop();
            k -= 1;
            if (k == 0) {
                return current.val;
            }
            
            current = current.right;
        }
        
        return -1; // Should not reach here if k is valid
    }
}

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Problem Info

Description

Problem 230: Kth Smallest Element in a BST - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Search Tree Binary Tree

Companies

AgodaCiscoGoogleOracle

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:230

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize an empty stack and set the current node to the root of the BST.
2
Step 2: While the current node is not null, push the current node onto the stack and move to its left child. This process continues until the leftmost node is reached.
3
Step 3: If the stack is empty, the traversal is complete. Otherwise, pop a node from the stack.
4
Step 4: Decrement `k`. If `k` is 0, return the value of the popped node (the k-th smallest element).
5
Step 5: Set the current node to the right child of the popped node, and repeat steps 2-4.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because in the worst-case, we might need to visit every node in the tree before finding the k-th smallest. The space complexity is O(n) due to the stack, which, in a skewed tree, could hold all the nodes. If the tree is balanced, the stack space would be proportional to the height of the tree, resulting in O(log n) space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive In-order Traversal

Perform an in-order traversal recursively and keep a count. Return the node value when the count equals k.

Time: O(n)Space: O(n) due to recursion stack in the worst case (skewed tree). O(log n) in the average case (balanced tree)

Trade-offs: Recursive approach is more concise but may have slightly higher overhead due to function calls. Iterative approach might be preferred if stack overflow is a concern for very deep trees.

Using Morris Traversal

Morris traversal allows in-order traversal without using extra space for a stack or recursion. It modifies the tree temporarily by creating temporary links.

Time: O(n)Space: O(1)

Trade-offs: Morris traversal modifies the tree structure, so it's not suitable if the tree needs to remain unchanged. It's also more complex to implement.

✨

Key Insights

1Insight 1: An in-order traversal of a BST yields the nodes in ascending order (sorted order). This is the fundamental property we leverage.
2Insight 2: Iterative in-order traversal using a stack avoids recursion, potentially saving space in certain scenarios.
3Insight 3: We can stop the in-order traversal once we have found the k-th smallest element, avoiding processing the entire tree.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is null). The provided code handles this correctly because the `while` loop terminates immediately, and the `return -1` will never be hit as stated in the comments. The function will only returns -1 if k is greater than the number of nodes in the tree.
Edge case 2: k is larger than the number of nodes in the BST. The code will return -1 in this case, indicating that the k-th smallest element does not exist. However, the constraints specify `1 <= k <= n`, so this case should not occur according to the problem statement.
Edge case 3: k is 1 (finding the minimum element). The algorithm correctly identifies the leftmost node as the minimum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using pre-order or post-order traversal instead of in-order. These traversals will not yield nodes in sorted order.

✗Common mistake 2: Not handling the case where k is larger than the number of nodes. Although the constraints state that k should be valid, it's good practice to handle this scenario gracefully.

✗Common mistake 3: Incorrectly implementing the iterative in-order traversal with a stack, leading to incorrect ordering of nodes.

✗Common mistake 4: Assuming the tree is balanced, therefore leading to incorrect analysis of space complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Database indexing. BSTs are used in database indexing to efficiently retrieve data in a sorted manner. Finding the k-th smallest element can be useful for implementing pagination or ranking queries.

🏦Real-world scenario 2: Finding percentiles. In statistical analysis, finding the k-th smallest element (quantile) is a common operation. BSTs can be used to maintain a sorted representation of data and efficiently calculate percentiles.

🎮Real-world scenario 3: Order statistics. Operating systems and other software often need to find the k-th smallest or largest element in a collection of data for resource allocation or scheduling.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 21, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Kth Smallest Element in a BST"?

For this medium-level Tree problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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