What is the time complexity of Lowest Common Ancestor of a Binary Search Tree?

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

What data structures are used to solve Lowest Common Ancestor of a Binary Search Tree?

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 Lowest Common Ancestor of a Binary Search Tree?

Insight 1: The BST property is crucial for an efficient solution. It allows us to determine whether 'p' and 'q' are in the left or right subtree of a node without searching the entire tree. Insight 2: If both 'p' and 'q' are greater than the current node, their LCA must be in the right subtree. If both are less than the current node, their LCA must be in the left subtree. If the current node's value is between 'p' and 'q', then the current node is the LCA.

Which companies ask LeetCode 235. Lowest Common Ancestor of a Binary Search Tree in interviews?

This problem is frequently asked at Google, LinkedIn, Oracle, Samsung, X and 1 more companies.

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Lowest Common Ancestor of a Binary Search Tree - LeetCode 235 Solution

Lowest Common Ancestor of a Binary Search Tree - Complete Solution Guide

Lowest Common Ancestor of a Binary Search Tree is LeetCode problem 235, 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 a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST. According to the definition of LCA on Wikipedia : “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself ).” Example 1: Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes 2 and 8 is 6. Example 2: Input: root =

Detailed Explanation

The problem asks us to find the Lowest Common Ancestor (LCA) of two given nodes, 'p' and 'q', in a Binary Search Tree (BST). The LCA is defined as the lowest node in the tree that has both 'p' and 'q' as descendants (where a node can be a descendant of itself). A BST has the property that for any node, all nodes in its left subtree have values less than the node's value, and all nodes in its right subtree have values greater than the node's value.

Solution Approach

The provided solution uses an iterative approach to find the LCA. Starting from the root, it traverses the tree based on the values of 'p' and 'q' relative to the current node. If both 'p' and 'q' are greater than the current node's value, it moves to the right subtree. If both are less than the current node's value, it moves to the left subtree. If the current node's value is between 'p' and 'q', or if one of 'p' or 'q' is equal to the current node, then the current node is the LCA.

Step-by-Step Algorithm

Step 1: Initialize 'current' to the root of the BST.
Step 2: While 'current' is not null, compare the values of 'p' and 'q' to the value of 'current'.
Step 3: If both 'p.val' and 'q.val' are greater than 'current.val', move 'current' to the right child ('current = current.right').
Step 4: If both 'p.val' and 'q.val' are less than 'current.val', move 'current' to the left child ('current = current.left').
Step 5: Otherwise, 'current' is the LCA. Return 'current'.
Step 6: If the loop finishes without finding an LCA (which shouldn't happen in a valid BST as per problem constraints), return null.

Key Insights

Insight 1: The BST property is crucial for an efficient solution. It allows us to determine whether 'p' and 'q' are in the left or right subtree of a node without searching the entire tree.
Insight 2: If both 'p' and 'q' are greater than the current node, their LCA must be in the right subtree. If both are less than the current node, their LCA must be in the left subtree. If the current node's value is between 'p' and 'q', then the current node is the LCA.
Insight 3: The iterative approach leverages the BST properties to find the LCA by traversing the tree without recursion, leading to constant space complexity.

Complexity Analysis

Time Complexity: O(h)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Google, LinkedIn, Oracle, Samsung, X, Yandex.

Lowest Common Ancestor of a Binary Search Tree | LeetCode 235

Medium

Tree Depth-First Search Binary Search Tree Binary Tree

Time: O(h)Space: O(1)

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        TreeNode current = root;
        
        while (current != null) {
            if (p.val > current.val && q.val > current.val) {
                current = current.right;
            } else if (p.val < current.val && q.val < current.val) {
                current = current.left;
            } else {
                return current;
            }
        }
        
        return null; // Should not happen in a valid BST
    }
}

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Problem Info

Description

Problem 235: Lowest Common Ancestor of a Binary Search Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Search Tree Binary Tree

Companies

GoogleLinkedInOracleSamsungXYandex

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:235

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

6 steps

1
Step 1: Initialize 'current' to the root of the BST.
2
Step 2: While 'current' is not null, compare the values of 'p' and 'q' to the value of 'current'.
3
Step 3: If both 'p.val' and 'q.val' are greater than 'current.val', move 'current' to the right child ('current = current.right').
4
Step 4: If both 'p.val' and 'q.val' are less than 'current.val', move 'current' to the left child ('current = current.left').
5
Step 5: Otherwise, 'current' is the LCA. Return 'current'.
6
Step 6: If the loop finishes without finding an LCA (which shouldn't happen in a valid BST as per problem constraints), return null.

💡

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(h)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the number of nodes we visit during traversal. The space complexity is constant because we only use a few variables during the traversal, and no additional data structures are created that depend on the size of the input.

📖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 Approach

A recursive approach mirrors the iterative one, performing similar checks and recursively calling the function on the left or right subtree based on the relative values of 'p', 'q', and the current node.

Time: O(h), where h is the height of the tree.Space: O(h) in the worst case (skewed tree) due to the recursion stack. O(log n) in the average case (balanced tree).

Trade-offs: The recursive approach is conceptually similar but might be slightly less efficient due to function call overhead and the potential for stack overflow in very deep trees. The iterative approach is generally preferred due to its constant space complexity.

✨

Key Insights

1Insight 1: The BST property is crucial for an efficient solution. It allows us to determine whether 'p' and 'q' are in the left or right subtree of a node without searching the entire tree.
2Insight 2: If both 'p' and 'q' are greater than the current node, their LCA must be in the right subtree. If both are less than the current node, their LCA must be in the left subtree. If the current node's value is between 'p' and 'q', then the current node is the LCA.
3Insight 3: The iterative approach leverages the BST properties to find the LCA by traversing the tree without recursion, leading to constant space complexity.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: 'p' or 'q' is the root. The algorithm correctly identifies the root as the LCA in such scenarios because the 'else' condition in the while loop will be met.
Edge case 2: 'p' and 'q' are in different subtrees of the root. The algorithm correctly traverses the tree to the point where the paths to 'p' and 'q' diverge. The node at that point is the LCA.
Edge case 3: One of 'p' or 'q' is an ancestor of the other (e.g., Example 2). The algorithm will eventually reach the ancestor node and correctly identify it as the LCA.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the BST property. Many beginners might try to apply a generic tree traversal algorithm (like DFS or BFS) without leveraging the sorted nature of the BST, resulting in a less efficient solution.

✗Common mistake 2: Incorrectly handling the case where 'p' or 'q' is equal to the LCA. The problem statement specifies that a node can be a descendant of itself. The 'else' condition must handle this scenario correctly.

✗Common mistake 3: Assuming the tree is always balanced. The worst-case time complexity is O(n) for a skewed tree, so it's important to understand the implications of an unbalanced tree.

✅

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: File system navigation. Consider a file system where directories are organized in a tree-like structure. Finding the common ancestor directory of two files can be solved using a similar LCA concept.

🏦Real-world scenario 2: Organizational hierarchy. In an organization chart represented as a tree, finding the common manager of two employees can be viewed as an LCA problem.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 26, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

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 "Lowest Common Ancestor of a Binary Search Tree"?

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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