What is the time complexity of Leaf-Similar Trees?

The optimal solution has a time complexity of O(N + M).

What is the space complexity of Leaf-Similar Trees?

The space complexity is O(N + M).

What data structures are used to solve Leaf-Similar Trees?

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

What are the key insights for solving Leaf-Similar Trees?

Insight 1: We need to extract the leaf node values in the correct order (left to right) for each tree. Insight 2: Depth-First Search (DFS) or Breadth-First Search (BFS) can be used to traverse the trees and identify the leaf nodes.

Which companies ask LeetCode 872. Leaf-Similar Trees in interviews?

This problem is frequently asked at Snowflake.

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Leaf-Similar Trees - LeetCode 872 Solution

Leaf-Similar Trees - Complete Solution Guide

Leaf-Similar Trees is LeetCode problem 872, 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

Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence . For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8) . Two binary trees are considered leaf-similar if their leaf value sequence is the same. Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar. Example 1: Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null

Detailed Explanation

The problem asks us to determine if two binary trees are 'leaf-similar'. Two trees are leaf-similar if their leaf value sequences are the same. A leaf value sequence is created by traversing the tree from left to right and listing the values of all the leaf nodes (nodes with no children). The input consists of the root nodes of the two binary trees, `root1` and `root2`. The output is a boolean value: `true` if the trees are leaf-similar, and `false` otherwise. The number of nodes in each tree is between 1 and 200, and the values of the nodes are between 0 and 200.

Solution Approach

The solution involves the following steps: 1) Write a function to traverse a binary tree and extract its leaf node values into a list, preserving the left-to-right order. Depth-First Search is a natural and efficient way to do this. 2) Apply this function to both input trees, `root1` and `root2`, to obtain their respective leaf value sequences. 3) Compare the two leaf value sequences. If they are identical, return `true`; otherwise, return `false`.

Step-by-Step Algorithm

Step 1: Define a recursive function `getLeafValues(root)` or an iterative equivalent to traverse the tree.
Step 2: Within the `getLeafValues` function, check if the current node is a leaf node (both left and right children are null).
Step 3: If the current node is a leaf node, add its value to the leaf value sequence (list).
Step 4: If the current node is not a leaf node, recursively call `getLeafValues` on its left child, and then on its right child. This ensures the left-to-right order of leaf nodes.
Step 5: In the main function `leafSimilar(root1, root2)`, call `getLeafValues` for both `root1` and `root2` to get two lists of leaf values.
Step 6: Compare the two lists for equality using `.equals()` in Java or `==` in Python/C++. Return `true` if they are equal, `false` otherwise.

Key Insights

Insight 1: We need to extract the leaf node values in the correct order (left to right) for each tree.
Insight 2: Depth-First Search (DFS) or Breadth-First Search (BFS) can be used to traverse the trees and identify the leaf nodes.
Insight 3: After extracting the leaf node sequences, we need to compare them for equality.

Complexity Analysis

Time Complexity: O(N + M)

Space Complexity: O(N + M)

Topics

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

Companies

Asked at: Snowflake.

Leaf-Similar Trees | LeetCode 872

Easy

Tree Depth-First Search Binary Tree

Time: O(N + M)Space: O(N + M)

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 leafSimilar(TreeNode root1, TreeNode root2) {
        List<Integer> leaves1 = new ArrayList<>();
        List<Integer> leaves2 = new ArrayList<>();
        dfs(root1, leaves1);
        dfs(root2, leaves2);
        return leaves1.equals(leaves2);
    }

    private void dfs(TreeNode node, List<Integer> leaves) {
        if (node == null) return;
        if (node.left == null && node.right == null) {
            leaves.add(node.val);
            return;
        }
        dfs(node.left, leaves);
        dfs(node.right, leaves);
    }
}

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Problem Info

Description

Problem 872: Leaf-Similar Trees - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

Snowflake

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:872

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: Define a recursive function `getLeafValues(root)` or an iterative equivalent to traverse the tree.
2
Step 2: Within the `getLeafValues` function, check if the current node is a leaf node (both left and right children are null).
3
Step 3: If the current node is a leaf node, add its value to the leaf value sequence (list).
4
Step 4: If the current node is not a leaf node, recursively call `getLeafValues` on its left child, and then on its right child. This ensures the left-to-right order of leaf nodes.
5
Step 5: In the main function `leafSimilar(root1, root2)`, call `getLeafValues` for both `root1` and `root2` to get two lists of leaf values.
6
Step 6: Compare the two lists for equality using `.equals()` in Java or `==` in Python/C++. Return `true` if they are equal, `false` otherwise.

💡

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 + M)

💾

Space Complexity

O(N + M)

📝

Why?

The time complexity is determined by the tree traversals, which visit each node once. The space complexity is determined by the size of the leaf lists, which, in the worst case (e.g., a skewed tree where every node is a leaf), can be proportional to the number of nodes in 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

Iterative Depth-First Search

Instead of recursion, an iterative DFS using a stack can be employed to traverse the tree. This avoids the overhead of recursive function calls and can be slightly more efficient in practice, although the asymptotic complexity remains the same.

Time: O(N + M)Space: O(N + M)

Trade-offs: Iterative DFS avoids recursion overhead. Useful when concerned about stack overflow issues with very deep trees. Code may be slightly more complex to read.

✨

Key Insights

1Insight 1: We need to extract the leaf node values in the correct order (left to right) for each tree.
2Insight 2: Depth-First Search (DFS) or Breadth-First Search (BFS) can be used to traverse the trees and identify the leaf nodes.
3Insight 3: After extracting the leaf node sequences, we need to compare them for equality.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree(s). If a tree is empty, its leaf value sequence is empty. The code correctly handles this because it returns an empty list in the `getLeafValues` function if the root is `None` or `null`.
Edge case 2: Single node tree. If a tree contains only the root node and no children, that node is a leaf. The code correctly identifies this and adds the root value to the leaf sequence.
Edge case 3: One tree is much larger than the other. The code still processes each tree independently and then compares the resulting lists. This does not cause any issues.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling empty trees correctly. Failing to account for null or empty trees can lead to errors.

✗Common mistake 2: Incorrectly identifying leaf nodes. A node is a leaf only if it has *no* left and right children.

✗Common mistake 3: Not preserving the left-to-right order of leaf nodes. This is crucial for determining if the trees are leaf-similar.

✗Common mistake 4: Assuming that if the root nodes are the same, the trees are leaf-similar. This is incorrect.

✅

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: Comparing the structure of XML or JSON documents. The leaf nodes can represent data values, and comparing leaf sequences can determine if two documents have the same data content.

🏦Real-world scenario 2: Analyzing phylogenetic trees in biology. Leaf nodes represent species, and comparing leaf sequences can help determine the similarity between evolutionary relationships.

💡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.

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 31, 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

❓

Frequently Asked Questions

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Leaf-Similar Trees"?

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.

🤖 Need help with this problem?