What is the time complexity of Univalued Binary Tree?

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

What data structures are used to solve Univalued Binary Tree?

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

What are the key insights for solving Univalued Binary Tree?

Insight 1: We can traverse the tree using Depth-First Search (DFS) to check each node's value. Insight 2: We need to compare each node's value with the value of the root node. If any node has a different value, the tree is not uni-valued.

Which companies ask LeetCode 965. Univalued Binary Tree in interviews?

This problem is frequently asked at Twilio.

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Univalued Binary Tree - LeetCode 965 Solution

Q: How to solve Univalued Binary Tree?

The solution uses a recursive Depth-First Search (DFS) approach. We start by storing the value of the root node. Then, we recursively traverse the tree, comparing the value of each node with the stored root value. If we encounter a node with a different value, we immediately return `false`. If we reach a null node, we return `true`. The recursion continues until all nodes have been visited (or a node with a different value is found).

Univalued Binary Tree - Complete Solution Guide

Univalued Binary Tree is LeetCode problem 965, 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

A binary tree is uni-valued if every node in the tree has the same value. Given the root of a binary tree, return true if the given tree is uni-valued , or false otherwise. Example 1: Input: root = [1,1,1,1,1,null,1] Output: true Example 2: Input: root = [2,2,2,5,2] Output: false Constraints: The number of nodes in the tree is in the range [1, 100] . 0 <= Node.val < 100

Detailed Explanation

The problem asks us to determine if a given binary tree is 'uni-valued'. A binary tree is uni-valued if every node in the tree has the same value. The input is the root of a binary tree, and the output should be a boolean: `true` if the tree is uni-valued, and `false` otherwise. The constraints specify the number of nodes is between 1 and 100, and node values are between 0 and 99.

Solution Approach

The solution uses a recursive Depth-First Search (DFS) approach. We start by storing the value of the root node. Then, we recursively traverse the tree, comparing the value of each node with the stored root value. If we encounter a node with a different value, we immediately return `false`. If we reach a null node, we return `true`. The recursion continues until all nodes have been visited (or a node with a different value is found).

Step-by-Step Algorithm

Step 1: Check if the root is null. If it is, return true (an empty tree is considered uni-valued).
Step 2: Store the value of the root node in a variable (e.g., `val`).
Step 3: Define a recursive helper function `is_unival(node)` or `isUnivalTreeHelper(node, val)` that takes a node as input.
Step 4: Inside the helper function, check if the node is null. If it is, return `true`.
Step 5: If the node is not null, compare its value with the stored root value (`val`).
Step 6: If the node's value is different from `val`, return `false`.
Step 7: If the node's value is the same as `val`, recursively call the helper function for the left and right subtrees: `return is_unival(node.left) and is_unival(node.right)`.
Step 8: Call the helper function with the root node initially: `return is_unival(root)`.

Key Insights

Insight 1: We can traverse the tree using Depth-First Search (DFS) to check each node's value.
Insight 2: We need to compare each node's value with the value of the root node. If any node has a different value, the tree is not uni-valued.
Insight 3: The base case for the recursive DFS is when we reach a null node. In that case, we return `true` because a null node doesn't violate the uni-valued property.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: Twilio.

Univalued Binary Tree | LeetCode 965

Easy

Tree Depth-First Search Breadth-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 {
    public boolean isUnivalTree(TreeNode root) {
        if (root == null) {
            return true;
        }
        return isUnivalTreeHelper(root, root.val);
    }

    private boolean isUnivalTreeHelper(TreeNode node, int value) {
        if (node == null) {
            return true;
        }
        if (node.val != value) {
            return false;
        }
        return isUnivalTreeHelper(node.left, value) && isUnivalTreeHelper(node.right, value);
    }
}

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Problem Info

Description

Problem 965: Univalued Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

Twilio

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:965

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

8 steps

1
Step 1: Check if the root is null. If it is, return true (an empty tree is considered uni-valued).
2
Step 2: Store the value of the root node in a variable (e.g., `val`).
3
Step 3: Define a recursive helper function `is_unival(node)` or `isUnivalTreeHelper(node, val)` that takes a node as input.
4
Step 4: Inside the helper function, check if the node is null. If it is, return `true`.
5
Step 5: If the node is not null, compare its value with the stored root value (`val`).
6
Step 6: If the node's value is different from `val`, return `false`.
7
Step 7: If the node's value is the same as `val`, recursively call the helper function for the left and right subtrees: `return is_unival(node.left) and is_unival(node.right)`.
8
Step 8: Call the helper function with the root node initially: `return is_unival(root)`.

💡

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 we visit each node at most once. The space complexity is O(h) due to the depth of the recursion stack, which depends on 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

Breadth-First Search (BFS)

We could use BFS instead of DFS. We would enqueue the root node, then iterate through the queue, checking if each node's value matches the root's value. If not, we return `false`. We enqueue the children of each node if they exist.

Time: O(n)Space: O(w), where w is the maximum width of the tree. In the worst case (a complete binary tree), w is approximately n/2, so the space complexity can be considered O(n).

Trade-offs: BFS might use more memory than DFS if the tree is skewed because BFS explores level by level and might store more nodes in the queue at once. DFS uses space proportional to the height of the tree, which could be smaller for balanced trees.

✨

Key Insights

1Insight 1: We can traverse the tree using Depth-First Search (DFS) to check each node's value.
2Insight 2: We need to compare each node's value with the value of the root node. If any node has a different value, the tree is not uni-valued.
3Insight 3: The base case for the recursive DFS is when we reach a null node. In that case, we return `true` because a null node doesn't violate the uni-valued property.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is null). The solution correctly handles this by returning `true` immediately.
Edge case 2: Single node tree. The solution also handles this correctly as the recursive calls immediately hit the null base case returning true.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the null root case. Failing to check for a null root will lead to a null pointer exception.

✗Common mistake 2: Forgetting to recursively check both the left and right subtrees. Only checking one subtree will result in incorrect results if the other subtree contains a different value.

✗Common mistake 3: Not comparing node values with the root value, but instead only comparing children values with their parents value. This approach fails on certain complex trees that are uni-valued.

✅

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: Verifying data integrity in a hierarchical data structure. If you have a tree-like data structure where all nodes should have a specific property, this algorithm can be used to check if the integrity of the data has been compromised.

🏦Real-world scenario 2: Configuration management. In a configuration management system where certain settings should be consistent across a hierarchy of objects, this algorithm can be used to verify that the configuration is applied correctly.

💡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

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

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Tree. It's a great foundation for more complex problems.

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: August 5, 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 14

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

@AlgoMasterJul 7

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

@InterviewProJun 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.

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 "Univalued Binary Tree"?

For this easy-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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