What is the time complexity of Two Sum IV - Input is a BST?

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

What data structures are used to solve Two Sum IV - Input is a BST?

This problem uses Hash Table, Two Pointers, Tree, Depth-First Search, Breadth-First Search, Binary Search Tree, Binary Tree. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Two Sum IV - Input is a BST?

Insight 1: The problem leverages the properties of a BST: values in the left subtree are smaller than the node's value, and values in the right subtree are larger. Insight 2: Using a hash set (or set) allows for efficient O(1) average time complexity lookups to check if the complement (k - node.val) exists in the tree.

Which companies ask LeetCode 653. Two Sum IV - Input is a BST in interviews?

This problem is frequently asked at Cisco, Samsung.

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Two Sum IV - Input is a BST - LeetCode 653 Solution

Q: How to solve Two Sum IV - Input is a BST?

The provided solutions use a Depth-First Search (DFS) traversal of the BST in combination with a hash set. The DFS explores each node, and for each node, it checks if the complement (k - node.val) already exists in the hash set. If it does, we've found a pair that sums to k, and we return 'true'. Otherwise, we add the current node's value to the hash set and continue the DFS traversal in both the left and right subtrees.

Two Sum IV - Input is a BST - Complete Solution Guide

Two Sum IV - Input is a BST is LeetCode problem 653, a Easy 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 true if there exist two elements in the BST such that their sum is equal to k , or false otherwise . Example 1: Input: root = [5,3,6,2,4,null,7], k = 9 Output: true Example 2: Input: root = [5,3,6,2,4,null,7], k = 28 Output: false Constraints: The number of nodes in the tree is in the range [1, 10 4 ] . -10 4 <= Node.val <= 10 4 root is guaranteed to be a valid binary search tree. -10 5 <= k <= 10 5

Detailed Explanation

The problem asks us to determine if there exist two distinct nodes within a given Binary Search Tree (BST) whose values sum up to a target integer 'k'. The input is the root node of the BST and the target value 'k'. The output should be a boolean: 'true' if such a pair exists, and 'false' otherwise. The constraints specify the size of the tree, the range of node values, and the range of the target value.

Solution Approach

The provided solutions use a Depth-First Search (DFS) traversal of the BST in combination with a hash set. The DFS explores each node, and for each node, it checks if the complement (k - node.val) already exists in the hash set. If it does, we've found a pair that sums to k, and we return 'true'. Otherwise, we add the current node's value to the hash set and continue the DFS traversal in both the left and right subtrees.

Step-by-Step Algorithm

Step 1: Initialize a hash set (e.g., 'nums' in Python, 'seen' in Java and C++) to store the values of visited nodes.
Step 2: Define a recursive function 'dfs(node)' that takes a node as input.
Step 3: Inside 'dfs(node)', check if the node is null. If so, return 'false' because we've reached the end of a branch without finding a suitable pair.
Step 4: Check if 'k - node.val' is present in the hash set. If it is, return 'true' because we've found a pair that sums to k.
Step 5: Add 'node.val' to the hash set.
Step 6: Recursively call 'dfs(node.left)' and 'dfs(node.right)' and return 'true' if either of these calls returns 'true'. This is because we only need to find one pair to satisfy the condition. If both return false then return false.
Step 7: Call 'dfs(root)' to start the traversal from the root node and return the result.

Key Insights

Insight 1: The problem leverages the properties of a BST: values in the left subtree are smaller than the node's value, and values in the right subtree are larger.
Insight 2: Using a hash set (or set) allows for efficient O(1) average time complexity lookups to check if the complement (k - node.val) exists in the tree.
Insight 3: The problem can also be solved using an in-order traversal to store the BST elements into a sorted array, followed by a two-pointer approach for finding the sum.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Two Pointers, Tree, Depth-First Search, Breadth-First Search, Binary Search Tree, Binary Tree.

Companies

Asked at: Cisco, Samsung.

Two Sum IV - Input is a BST | LeetCode 653

Easy

Hash Table Two Pointers Tree Depth-First Search Breadth-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 {
    Set<Integer> seen = new HashSet<>();
    public boolean findTarget(TreeNode root, int k) {
        if (root == null) return false;
        if (seen.contains(k - root.val)) return true;
        seen.add(root.val);
        return findTarget(root.left, k) || findTarget(root.right, k);
    }
}

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Problem Info

Description

Problem 653: Two Sum IV - Input is a BST - Detailed explanation and solution approaches.

Topics

Hash Table Two Pointers Tree Depth-First Search Breadth-First Search Binary Search Tree Binary Tree

Companies

CiscoSamsung

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:653

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

7 steps

1
Step 1: Initialize a hash set (e.g., 'nums' in Python, 'seen' in Java and C++) to store the values of visited nodes.
2
Step 2: Define a recursive function 'dfs(node)' that takes a node as input.
3
Step 3: Inside 'dfs(node)', check if the node is null. If so, return 'false' because we've reached the end of a branch without finding a suitable pair.
4
Step 4: Check if 'k - node.val' is present in the hash set. If it is, return 'true' because we've found a pair that sums to k.
5
Step 5: Add 'node.val' to the hash set.
6
Step 6: Recursively call 'dfs(node.left)' and 'dfs(node.right)' and return 'true' if either of these calls returns 'true'. This is because we only need to find one pair to satisfy the condition. If both return false then return false.
7
Step 7: Call 'dfs(root)' to start the traversal from the root node and return the result.

💡

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 determined by the number of nodes visited during the DFS traversal, which is at most 'n'. The space complexity is mainly determined by the size of the hash set, which can store up to 'n' elements in the worst case. In the case of recursion it can be the height of the tree.

📖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

In-order Traversal and Two Pointers

Perform an in-order traversal of the BST to store the nodes' values in a sorted array. Then, use the two-pointer technique on the sorted array to find two elements that sum to 'k'.

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

Trade-offs: This approach is an alternative but potentially less efficient in terms of space if the tree is relatively balanced, as the hash set might not grow to 'n' in such cases in the main solution. However, if memory allocation is cheap, this two-pointer approach might be faster in many practical cases.

✨

Key Insights

1Insight 1: The problem leverages the properties of a BST: values in the left subtree are smaller than the node's value, and values in the right subtree are larger.
2Insight 2: Using a hash set (or set) allows for efficient O(1) average time complexity lookups to check if the complement (k - node.val) exists in the tree.
3Insight 3: The problem can also be solved using an in-order traversal to store the BST elements into a sorted array, followed by a two-pointer approach for finding the sum.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: The root is null. The code handles this correctly by returning 'false' because an empty tree cannot contain a pair that sums to 'k'.
Edge case 2: The BST contains duplicate values. The hash set will only store one instance of the duplicate value, potentially leading to incorrect results if the two values that sum to k are the same duplicate value. However, the problem states that we need *two elements*. If k = 2*x and x exists more than once in the tree, then an alternative solution where we collect the BST into an array and then apply a two-pointer approach can be easily modified to account for repeated value. In this DFS with HashSet approach the duplicated value x is only stored once in the seen set, so only values differing from x will be searched for. Since our search starts with x, and continues with other values, once we stumble on another node y such that k - y == x, we found the answer.
Edge case 3: No two elements sum up to k. The algorithm correctly traverses the entire tree without finding a match, and returns 'false'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the root is null. This will lead to a NullPointerException or incorrect results.

✗Common mistake 2: Using a data structure other than a hash set or hash map for lookups. This will result in a time complexity greater than O(1) for lookups, potentially increasing the overall time complexity.

✗Common mistake 3: Forgetting to consider the properties of BST - the algorithm doesn't depend on the values being sorted, so a general binary tree could be used with this algorithm.

✅

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 query optimization. Imagine a database structured as a BST. You might use this algorithm to quickly determine if there are two entries with a combined value equal to a specific target for reporting purposes.

🏦Real-world scenario 2: Financial analysis. You could use this algorithm to determine if there are two assets in a portfolio whose combined value meets a specific investment threshold.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: September 22, 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 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Two Sum IV - Input is a BST"?

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

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