What is the time complexity of Longest Univalue Path?

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

What data structures are used to solve Longest Univalue Path?

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

What are the key insights for solving Longest Univalue Path?

Insight 1: The longest univalue path at a node can either be contained within the left subtree, the right subtree, or pass through the current node. Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the length of the univalue path at each node.

Which companies ask LeetCode 687. Longest Univalue Path in interviews?

This problem is frequently asked at Snowflake, Sprinklr, Zepto.

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Longest Univalue Path - LeetCode 687 Solution

Longest Univalue Path - Complete Solution Guide

Longest Univalue Path is LeetCode problem 687, 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 tree, return the length of the longest path, where each node in the path has the same value . This path may or may not pass through the root. The length of the path between two nodes is represented by the number of edges between them. Example 1: Input: root = [5,4,5,1,1,null,5] Output: 2 Explanation: The shown image shows that the longest path of the same value (i.e. 5). Example 2: Input: root = [1,4,5,4,4,null,5] Output: 2 Explanation: The shown image shows that the l

Detailed Explanation

The problem asks us to find the length of the longest path in a binary tree where all the nodes in the path have the same value. The path may or may not pass through the root. The length of a path is defined as the number of edges between the nodes.

Solution Approach

The provided solution uses a recursive Depth-First Search (DFS) approach. The `dfs` function calculates the length of the longest univalue path that starts at a given node and extends downwards. It recursively computes the longest univalue paths from the left and right children. If the left (or right) child has the same value as the current node, we extend the path; otherwise, we start a new path. The global variable `ans` keeps track of the longest univalue path found so far, irrespective of whether it includes the current node.

Step-by-Step Algorithm

Step 1: Initialize a global variable `ans` to 0. This variable will store the length of the longest univalue path found so far.
Step 2: Define a recursive function `dfs(node)` that takes a node as input.
Step 3: Base Case: If the node is null, return 0.
Step 4: Recursively call `dfs` on the left and right children of the node to get the lengths of the longest univalue paths starting at those children (left_length and right_length).
Step 5: Initialize `left_arrow` and `right_arrow` to 0. These variables will store the lengths of the longest univalue paths extending from the current node to the left and right, respectively.
Step 6: If the left child exists and has the same value as the current node, set `left_arrow` to `left_length + 1`. Otherwise, `left_arrow` remains 0, indicating no extension to the left.
Step 7: Similarly, if the right child exists and has the same value as the current node, set `right_arrow` to `right_length + 1`. Otherwise, `right_arrow` remains 0, indicating no extension to the right.
Step 8: Update the global variable `ans` to the maximum of its current value and the sum of `left_arrow` and `right_arrow`. This step ensures that we keep track of the longest univalue path found so far, even if it passes through the current node.
Step 9: Return the maximum of `left_arrow` and `right_arrow`. This value represents the length of the longest univalue path that starts at the current node and extends downwards to either the left or the right.
Step 10: Call `dfs(root)` to start the traversal from the root of the tree.
Step 11: Return the value of `ans`, which represents the length of the longest univalue path in the tree.

Key Insights

Insight 1: The longest univalue path at a node can either be contained within the left subtree, the right subtree, or pass through the current node.
Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the length of the univalue path at each node.
Insight 3: The path length is measured by the number of *edges* between nodes, which is one less than the number of nodes if counting nodes in the path. The code returns the edge count.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: Snowflake, Sprinklr, Zepto.

Longest Univalue Path | LeetCode 687

Medium

Tree Depth-First Search Binary Tree

Time: O(n)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;
 *     }
 * }
 */
class Solution {
    int ans = 0;
    public int longestUnivaluePath(TreeNode root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode node) {
        if (node == null) {
            return 0;
        }

        int left_length = dfs(node.left);
        int right_length = dfs(node.right);

        int left_arrow = 0;
        if (node.left != null && node.left.val == node.val) {
            left_arrow = left_length + 1;
        }

        int right_arrow = 0;
        if (node.right != null && node.right.val == node.val) {
            right_arrow = right_length + 1;
        }

        ans = Math.max(ans, left_arrow + right_arrow);

        return Math.max(left_arrow, right_arrow);
    }
}

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Problem Info

Description

Problem 687: Longest Univalue Path - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

SnowflakeSprinklrZepto

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:687

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

11 steps

1
Step 1: Initialize a global variable `ans` to 0. This variable will store the length of the longest univalue path found so far.
2
Step 2: Define a recursive function `dfs(node)` that takes a node as input.
3
Step 3: Base Case: If the node is null, return 0.
4
Step 4: Recursively call `dfs` on the left and right children of the node to get the lengths of the longest univalue paths starting at those children (left_length and right_length).
5
Step 5: Initialize `left_arrow` and `right_arrow` to 0. These variables will store the lengths of the longest univalue paths extending from the current node to the left and right, respectively.
6
Step 6: If the left child exists and has the same value as the current node, set `left_arrow` to `left_length + 1`. Otherwise, `left_arrow` remains 0, indicating no extension to the left.
7
Step 7: Similarly, if the right child exists and has the same value as the current node, set `right_arrow` to `right_length + 1`. Otherwise, `right_arrow` remains 0, indicating no extension to the right.
8
Step 8: Update the global variable `ans` to the maximum of its current value and the sum of `left_arrow` and `right_arrow`. This step ensures that we keep track of the longest univalue path found so far, even if it passes through the current node.
9
Step 9: Return the maximum of `left_arrow` and `right_arrow`. This value represents the length of the longest univalue path that starts at the current node and extends downwards to either the left or the right.
10
Step 10: Call `dfs(root)` to start the traversal from the root of the tree.
11
Step 11: Return the value of `ans`, which represents the length of the longest univalue path in the tree.

💡

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(h)

📝

Why?

The time complexity is O(n) because the DFS algorithm visits each node in the binary tree once. The space complexity is determined by the depth of the recursion stack, which corresponds to the height of the tree. In a balanced tree, the height is logarithmic, whereas in a skewed tree, it's linear.

📖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

Iterative DFS

An iterative DFS approach using a stack could be used instead of recursion. This would eliminate the risk of stack overflow for very deep trees.

Time: O(n)Space: O(n) in the worst case (skewed tree), O(w) on average where w is the maximum width of the tree

Trade-offs: Iterative DFS might be slightly more complex to implement, but it avoids the recursion overhead and potential stack overflow issues for very deep trees. The space complexity might be O(w) in average case if breadth first order is followed while pushing children nodes to the stack.

✨

Key Insights

1Insight 1: The longest univalue path at a node can either be contained within the left subtree, the right subtree, or pass through the current node.
2Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the length of the univalue path at each node.
3Insight 3: The path length is measured by the number of *edges* between nodes, which is one less than the number of nodes if counting nodes in the path. The code returns the edge count.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty Tree: If the input tree is empty (root is None), the code correctly handles this case by returning 0 since the dfs function immediately returns 0.
Edge case 2: Tree with a single node: If the tree has only one node, the dfs function returns 0, and ans remains initialized to 0, resulting in the correct output.
Edge case 3: All nodes have different values: The algorithm correctly identifies that the longest univalue path in this case is 0, as no two adjacent nodes have the same value.
Edge case 4: Skewed Tree: The algorithm handles skewed trees correctly. The 'ans' variable will get updated during DFS and will eventually hold the length of the longest univalue path.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering paths that do not pass through the root. The problem explicitly states the path does not have to pass through the root, so the global `ans` is vital.

✗Common mistake 2: Incorrectly calculating the path length. The path length is the number of *edges* between the nodes, not the number of nodes in the path. Subtract 1 to convert nodes to edges if necessary.

✗Common mistake 3: Failing to reset the length of the path when the node values do not match. The recursive approach resets the path lengths by returning 0 when a value mismatch occurs.

✅

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: Analyzing network topologies: Identifying the longest chain of computers within a network that are running the same operating system or using the same protocol.

🏦Real-world scenario 2: Bioinformatics: Finding the longest sequence of identical amino acids in a protein structure, which could indicate a specific functional domain.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

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: October 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

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 "Longest Univalue Path"?

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.

🤖 Need help with this problem?