What is the time complexity of Binary Search Tree Iterator?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Binary Search Tree Iterator?

This problem uses Stack, Tree, Design, Binary Search Tree, Binary Tree, Iterator. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Binary Search Tree Iterator?

Insight 1: Using a stack to simulate the in-order traversal of the BST is crucial for achieving the desired time and space complexity. Insight 2: The `next()` operation must maintain the in-order traversal sequence by pushing the left path of the right subtree (if it exists) onto the stack.

Which companies ask LeetCode 173. Binary Search Tree Iterator in interviews?

This problem is frequently asked at LinkedIn, Media.net.

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Binary Search Tree Iterator - LeetCode 173 Solution

Binary Search Tree Iterator - Complete Solution Guide

Binary Search Tree Iterator is LeetCode problem 173, 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

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST): BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST. boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false . int next() Moves the poin

Detailed Explanation

The problem requires implementing an iterator for a Binary Search Tree (BST) that returns elements in in-order traversal sequence. The `BSTIterator` class needs to provide three methods: `BSTIterator(TreeNode root)` to initialize the iterator with the root of the BST, `next()` to return the next smallest element in the BST, and `hasNext()` to check if there are more elements available. The key constraint is to implement `next()` and `hasNext()` in average O(1) time and O(h) space, where 'h' is the height of the tree.

Solution Approach

The provided solution uses a stack to keep track of the nodes that need to be visited during the in-order traversal. The constructor initializes the stack by pushing the leftmost path of the BST onto the stack. The `next()` method pops a node from the stack (which is the next smallest element), and then pushes the leftmost path of the right subtree of that node onto the stack. The `hasNext()` method simply checks if the stack is empty. This approach amortizes the cost of traversing the tree to achieve O(1) average time complexity for both `next()` and `hasNext()` operations.

Step-by-Step Algorithm

Step 1: **Initialization (Constructor):** Given the root, push the root and all of its left descendants onto the stack. This sets up the iterator to start at the smallest element.
Step 2: **next() operation:** Pop the top element from the stack. This is the next smallest element in the BST.
Step 3: If the popped node has a right child, push the right child and all of its left descendants onto the stack.
Step 4: Return the value of the popped node.
Step 5: **hasNext() operation:** Check if the stack is empty. If it's not empty, there are more elements to visit; otherwise, the traversal is complete.

Key Insights

Insight 1: Using a stack to simulate the in-order traversal of the BST is crucial for achieving the desired time and space complexity.
Insight 2: The `next()` operation must maintain the in-order traversal sequence by pushing the left path of the right subtree (if it exists) onto the stack.
Insight 3: The stack will contain nodes that haven't been visited yet, maintaining the state of the traversal.
Insight 4: Pre-calculating the entire in-order traversal and storing it in a list would violate the O(h) space requirement since it can be O(n) in the worst case

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(H)

Topics

This problem involves: Stack, Tree, Design, Binary Search Tree, Binary Tree, Iterator.

Companies

Asked at: LinkedIn, Media.net.

Binary Search Tree Iterator | LeetCode 173

Medium

Stack Tree Design Binary Search Tree Binary Tree Iterator

Time: O(1)Space: O(H)

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;
 *     }
 * }
 */
import java.util.Stack;

class BSTIterator {

    private Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        _push_left_path(root);
    }
    
    public int next() {
        TreeNode node = stack.pop();
        if (node.right != null) {
            _push_left_path(node.right);
        }
        return node.val;
    }
    
    public boolean hasNext() {
        return !stack.isEmpty();
    }

    private void _push_left_path(TreeNode node) {
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

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Problem Info

Description

Problem 173: Binary Search Tree Iterator - Detailed explanation and solution approaches.

Topics

Stack Tree Design Binary Search Tree Binary Tree Iterator

Companies

LinkedInMedia.net

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:173

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: **Initialization (Constructor):** Given the root, push the root and all of its left descendants onto the stack. This sets up the iterator to start at the smallest element.
2
Step 2: **next() operation:** Pop the top element from the stack. This is the next smallest element in the BST.
3
Step 3: If the popped node has a right child, push the right child and all of its left descendants onto the stack.
4
Step 4: Return the value of the popped node.
5
Step 5: **hasNext() operation:** Check if the stack is empty. If it's not empty, there are more elements to visit; otherwise, the traversal is complete.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(H)

📝

Why?

The time complexity is amortized O(1) because the push_left_path operation, which can take O(h) time in the worst case (skewed tree), is not called in every next() operation. It's only called when the node popped from the stack has a right child. Therefore, the total time complexity for m next() operations is O(m+n) where n is the total number of nodes in the tree, as each node is visited and pushed onto the stack at most once. Thus the average cost per next() call is O((m+n)/m) which is O(1) since m >= n. The space complexity is O(h) as the height of the tree is the maximum size the stack can achieve.

📖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

Precompute In-order Traversal

Perform an in-order traversal of the BST and store the nodes in a list. The iterator then maintains an index into this list.

Time: O(n) for initialization, O(1) for next() and hasNext()Space: O(n)

Trade-offs: Simple to implement, but uses O(n) space, which violates the problem constraint of O(h) space.

✨

Key Insights

1Insight 1: Using a stack to simulate the in-order traversal of the BST is crucial for achieving the desired time and space complexity.
2Insight 2: The `next()` operation must maintain the in-order traversal sequence by pushing the left path of the right subtree (if it exists) onto the stack.
3Insight 3: The stack will contain nodes that haven't been visited yet, maintaining the state of the traversal.
4Insight 4: Pre-calculating the entire in-order traversal and storing it in a list would violate the O(h) space requirement since it can be O(n) in the worst case

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree. The iterator should handle an empty tree gracefully. The constructor should initialize an empty stack. The `hasNext()` method should return false immediately, and `next()` should not be called.
Edge case 2: Skewed tree (all left or all right). The stack will contain all nodes in the left skewed tree, leading to O(n) space complexity, where n is number of nodes in tree. However, the solution still works correctly.
Edge case 3: Root with only a right subtree. The code correctly handles pushing elements from the right subtree and its leftmost descendants.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the right subtree correctly in the `next()` operation. Forgetting to push the leftmost path of the right subtree onto the stack will lead to incorrect in-order traversal.

✗Common mistake 2: Using O(n) space by storing the entire in-order traversal in a list. The problem requires O(h) space.

✅

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 systems use iterators to traverse query results efficiently, and this BST iterator concept can be adapted to traverse sorted data structures.

🏦Real-world scenario 2: Compilers use iterators to process abstract syntax trees (ASTs) during code analysis and optimization. A BST iterator could be used if the AST is structured as a balanced tree.

💡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

Stacks are perfect for problems involving matching (parentheses) or "next greater element" patterns. If you need to look back at recent elements, think stack.

🎯 Interview Tip

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

⚠️ Watch Out

Popping from an empty stack. Always check if the stack is non-empty before popping.

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: November 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Binary Search Tree Iterator"?

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

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