What is the time complexity of Step-By-Step Directions From a Binary Tree Node to Another?

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

What data structures are used to solve Step-By-Step Directions From a Binary Tree Node to Another?

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

What are the key insights for solving Step-By-Step Directions From a Binary Tree Node to Another?

Insight 1: Finding the lowest common ancestor (LCA) implicitly is key. The paths from the root to `startValue` and `destValue` will share a common prefix that corresponds to the path from the root to their LCA. Insight 2: The path from `startValue` to `destValue` can be broken down into two segments: going 'up' from `startValue` to the LCA and then going 'down' from the LCA to `destValue`.

Which companies ask LeetCode 2096. Step-By-Step Directions From a Binary Tree Node to Another in interviews?

This problem is frequently asked at Databricks, Snowflake, TikTok.

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Step-By-Step Directions From a Binary Tree Node to Another - LeetCode 2096 Solution

Step-By-Step Directions From a Binary Tree Node to Another - Complete Solution Guide

Step-By-Step Directions From a Binary Tree Node to Another is LeetCode problem 2096, 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

You are given the root of a binary tree with n nodes. Each node is uniquely assigned a value from 1 to n . You are also given an integer startValue representing the value of the start node s , and a different integer destValue representing the value of the destination node t . Find the shortest path starting from node s and ending at node t . Generate step-by-step directions of such path as a string consisting of only the uppercase letters 'L' , 'R' , and 'U' . Each letter indicates a specific d

Detailed Explanation

The problem requires finding the shortest path between two nodes, `startValue` and `destValue`, in a binary tree and expressing this path as a string of 'U', 'L', and 'R' characters, representing moving up to the parent, left to the left child, and right to the right child, respectively. The tree's nodes have unique values. The goal is to find the sequence of moves that takes you from the `startValue` node to the `destValue` node using the fewest steps.

Solution Approach

The solution uses Depth-First Search (DFS) to find the paths from the root node to the `startValue` and `destValue` nodes. After finding the paths, it identifies the common prefix between them, which represents the path from the root to their lowest common ancestor (LCA). Then, it constructs the final path string. It first moves from the `startValue` node to the LCA by appending 'U' characters to the result string for each step. Then it goes from the LCA to the `destValue` node by appending the path from LCA to `destValue` to the string of 'U's.

Step-by-Step Algorithm

Step 1: Define a recursive DFS function `find_path(node, target, path)` to find the path from the root to a target node. The path is built as the recursion proceeds.
Step 2: Call the `find_path` function to find the path from the root to the `startValue` and store the path as a sequence of 'L' and 'R' characters.
Step 3: Call the `find_path` function to find the path from the root to the `destValue` and store the path as a sequence of 'L' and 'R' characters.
Step 4: Determine the length of the common prefix between the two paths found in steps 2 and 3.
Step 5: Construct the 'up' part of the path (from `startValue` to the LCA). Append 'U' to the result string (path to be returned) as many times as the difference between the length of the path from the root to `startValue` and the length of the common prefix.
Step 6: Construct the 'down' part of the path (from the LCA to `destValue`). Append the remaining part of the path from the root to `destValue` (i.e., the suffix after removing the common prefix) to the result string.
Step 7: Return the final result string representing the shortest path from `startValue` to `destValue`.

Key Insights

Insight 1: Finding the lowest common ancestor (LCA) implicitly is key. The paths from the root to `startValue` and `destValue` will share a common prefix that corresponds to the path from the root to their LCA.
Insight 2: The path from `startValue` to `destValue` can be broken down into two segments: going 'up' from `startValue` to the LCA and then going 'down' from the LCA to `destValue`.
Insight 3: Using Depth-First Search (DFS) to find the paths from the root to the `startValue` and `destValue` nodes is an efficient approach.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Databricks, Snowflake, TikTok.

Step-By-Step Directions From a Binary Tree Node to Another | LeetCode 2096

Medium

String Tree Depth-First Search 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;
 *     }
 * }
 */
import java.util.ArrayList;
import java.util.List;

class Solution {
    public String getDirections(TreeNode root, int startValue, int destValue) {
        
        List<Character> path_to_start = new ArrayList<>();
        find_path(root, startValue, path_to_start);
        
        List<Character> path_to_dest = new ArrayList<>();
        find_path(root, destValue, path_to_dest);
        
        int common_prefix_len = 0;
        while (common_prefix_len < path_to_start.size() &&
               common_prefix_len < path_to_dest.size() &&
               path_to_start.get(common_prefix_len).equals(path_to_dest.get(common_prefix_len))) {
            common_prefix_len += 1;
        }
        
        StringBuilder up_moves = new StringBuilder();
        for (int i = 0; i < path_to_start.size() - common_prefix_len; i++) {
            up_moves.append('U');
        }
        
        StringBuilder down_moves = new StringBuilder();
        for (int i = common_prefix_len; i < path_to_dest.size(); i++) {
            down_moves.append(path_to_dest.get(i));
        }
        
        return up_moves.toString() + down_moves.toString();
    }
    
    private boolean find_path(TreeNode node, int target, List<Character> path) {
        if (node == null) {
            return false;
        }
        
        if (node.val == target) {
            return true;
        }
        
        path.add('L');
        if (find_path(node.left, target, path)) {
            return true;
        }
        path.remove(path.size() - 1);
        
        path.add('R');
        if (find_path(node.right, target, path)) {
            return true;
        }
        path.remove(path.size() - 1);
        
        return false;
    }
}

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Problem Info

Description

Problem 2096: Step-By-Step Directions From a Binary Tree Node to Another - Detailed explanation and solution approaches.

Topics

String Tree Depth-First Search Binary Tree

Companies

DatabricksSnowflakeTikTok

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2096

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: Define a recursive DFS function `find_path(node, target, path)` to find the path from the root to a target node. The path is built as the recursion proceeds.
2
Step 2: Call the `find_path` function to find the path from the root to the `startValue` and store the path as a sequence of 'L' and 'R' characters.
3
Step 3: Call the `find_path` function to find the path from the root to the `destValue` and store the path as a sequence of 'L' and 'R' characters.
4
Step 4: Determine the length of the common prefix between the two paths found in steps 2 and 3.
5
Step 5: Construct the 'up' part of the path (from `startValue` to the LCA). Append 'U' to the result string (path to be returned) as many times as the difference between the length of the path from the root to `startValue` and the length of the common prefix.
6
Step 6: Construct the 'down' part of the path (from the LCA to `destValue`). Append the remaining part of the path from the root to `destValue` (i.e., the suffix after removing the common prefix) to the result string.
7
Step 7: Return the final result string representing the shortest path from `startValue` to `destValue`.

💡

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 we visit each node at most once during the DFS traversal. The space complexity is O(n) due to the recursion stack in DFS, as well as storing paths which could grow to the size of the tree's height in a worst-case skew 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

Lowest Common Ancestor (LCA) Approach

First, find the LCA of `startValue` and `destValue`. Then, find the path from `startValue` to the LCA and from the LCA to `destValue`. Combine these paths with appropriate 'U' characters.

Time: O(n) in the worst case, where n is the number of nodes in the tree.Space: O(n) in the worst case, where n is the number of nodes in the tree (for recursion stack and path storage).

Trade-offs: This approach is conceptually cleaner, but may involve more overhead in finding the LCA separately. The main solution implicitly calculates the LCA during the path finding process, making it slightly more efficient.

✨

Key Insights

1Insight 1: Finding the lowest common ancestor (LCA) implicitly is key. The paths from the root to `startValue` and `destValue` will share a common prefix that corresponds to the path from the root to their LCA.
2Insight 2: The path from `startValue` to `destValue` can be broken down into two segments: going 'up' from `startValue` to the LCA and then going 'down' from the LCA to `destValue`.
3Insight 3: Using Depth-First Search (DFS) to find the paths from the root to the `startValue` and `destValue` nodes is an efficient approach.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `startValue` is the parent of `destValue`. The path would simply be a single 'L' or 'R'. The solution handles this correctly because the common prefix length will be one less than length of path to destValue.
Edge case 2: `destValue` is the parent of `startValue`. The path will be a single 'U'. The solution handles this because the common prefix will be one less than the length of the path to the startValue.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where one node is an ancestor of the other correctly. The 'U' and 'L/R' path segments must be constructed based on the length of the common path prefix.

✗Common mistake 2: Incorrectly constructing the path from the LCA to the destination. Ensure you take only the *suffix* of the destination path after removing the common prefix, to avoid redundant steps.

✅

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: Pathfinding in a hierarchical organizational structure. For example, finding the steps to go from one employee to another in a company org chart.

🏦Real-world scenario 2: Navigation in a file system. Constructing a path string to navigate from one directory to another based on the hierarchical structure of the file system.

🎮Real-world scenario 3: Routing in computer networks. Representing the path from a source to a destination node in a network, where each node is connected in a tree-like structure.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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 25, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

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 "Step-By-Step Directions From a Binary Tree Node to Another"?

For this medium-level String 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.

🤖 Need help with this problem?