What is the time complexity of Binary Tree Paths?

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

What data structures are used to solve Binary Tree Paths?

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

What are the key insights for solving Binary Tree Paths?

Insight 1: A Depth-First Search (DFS) is the most efficient way to traverse the tree and explore all possible paths. Insight 2: Recursion is a natural fit for DFS on trees, simplifying the code and making it easier to understand.

Which companies ask LeetCode 257. Binary Tree Paths in interviews?

This problem is frequently asked at Capital One, Google, Revolut.

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Binary Tree Paths - LeetCode 257 Solution

Binary Tree Paths - Complete Solution Guide

Binary Tree Paths is LeetCode problem 257, 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

Given the root of a binary tree, return all root-to-leaf paths in any order . A leaf is a node with no children. Example 1: Input: root = [1,2,3,null,5] Output: ["1->2->5","1->3"] Example 2: Input: root = [1] Output: ["1"] Constraints: The number of nodes in the tree is in the range [1, 100] . -100 <= Node.val <= 100

Detailed Explanation

The problem asks you to find all possible paths from the root node to a leaf node in a binary tree. A path is a sequence of connected nodes from the root to a leaf (a node with no children). The output should be a list of strings, where each string represents a path and the node values are separated by "->". For example, if a path goes through nodes with values 1, 2, and 5, the corresponding string would be "1->2->5".

Solution Approach

The provided solutions all use a recursive Depth-First Search (DFS) approach. The function `dfs` (or its equivalent) recursively explores the tree. At each node, it appends the node's value to the current path string. If the current node is a leaf node (no left or right children), the path is added to the result list. Otherwise, the function recursively calls itself on the left and right children, adding "->" to the path string before each recursive call to separate node values.

Step-by-Step Algorithm

Step 1: Initialize an empty list to store the paths.
Step 2: Start a recursive DFS traversal from the root node.
Step 3: At each node, append the node's value to the current path string.
Step 4: If the current node is a leaf node, add the current path string to the result list.
Step 5: If the current node is not a leaf node, recursively call the DFS function for the left and right children, appending "->" to the path string before each recursive call.
Step 6: Return the list of paths.

Key Insights

Insight 1: A Depth-First Search (DFS) is the most efficient way to traverse the tree and explore all possible paths.
Insight 2: Recursion is a natural fit for DFS on trees, simplifying the code and making it easier to understand.
Insight 3: String concatenation within the recursive calls efficiently builds the path strings as we traverse down the tree. We only append the path to the result list when we reach a leaf node.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Capital One, Google, Revolut.

Binary Tree Paths | LeetCode 257

Easy

String Backtracking 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;
 *     }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new ArrayList<>();
        if (root == null) return paths;
        dfs(root, "", paths);
        return paths;
    }

    private void dfs(TreeNode node, String path, List<String> paths) {
        if (node.left == null && node.right == null) {
            paths.add(path + node.val);
            return;
        }
        path += node.val + "->";
        if (node.left != null) dfs(node.left, path, paths);
        if (node.right != null) dfs(node.right, path, paths);
    }
}

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Problem Info

Description

Problem 257: Binary Tree Paths - Detailed explanation and solution approaches.

Topics

String Backtracking Tree Depth-First Search Binary Tree

Companies

Capital OneGoogleRevolut

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:257

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 an empty list to store the paths.
2
Step 2: Start a recursive DFS traversal from the root node.
3
Step 3: At each node, append the node's value to the current path string.
4
Step 4: If the current node is a leaf node, add the current path string to the result list.
5
Step 5: If the current node is not a leaf node, recursively call the DFS function for the left and right children, appending "->" to the path string before each recursive call.
6
Step 6: Return the list of paths.

💡

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 linear because each node is visited and processed once. The space complexity is determined by the depth of the recursion (proportional to the tree's height) and the space used to store the resulting paths. In a balanced binary tree, the height is log(n), but in a skewed tree, the height can be n.

📖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 using a Stack

Instead of recursion, an iterative approach using a stack can be used to perform DFS. This avoids the risk of stack overflow for very deep trees.

Time: O(n)Space: O(n)

Trade-offs: The iterative approach might be slightly less concise than the recursive one but can handle extremely deep trees that might cause stack overflow with recursion.

Breadth-First Search (BFS)

While DFS is more natural for this problem, BFS could also be used. It would require a queue and a way to track the path for each node explored.

Time: O(n)Space: O(n)

Trade-offs: BFS would have similar time and space complexity but might be less efficient in practice due to the overhead of managing the queue.

✨

Key Insights

1Insight 1: A Depth-First Search (DFS) is the most efficient way to traverse the tree and explore all possible paths.
2Insight 2: Recursion is a natural fit for DFS on trees, simplifying the code and making it easier to understand.
3Insight 3: String concatenation within the recursive calls efficiently builds the path strings as we traverse down the tree. We only append the path to the result list when we reach a leaf node.

📚

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: Empty tree (root is null). The solutions correctly handle this by returning an empty list.
Edge case 2: Tree with only one node (root). The solutions correctly return a list containing only the root's value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling string concatenation. Forgetting to add "->" between node values or adding it at the wrong places can lead to incorrect path strings.

✗Common mistake 2: Not handling the base case (empty tree or leaf node) correctly. This can cause infinite recursion or incorrect output.

✅

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: Finding all possible routes in a network (represented as a tree). For example, finding all possible paths from a source node to a destination node in a communication network.

🏦Real-world scenario 2: Exploring hierarchical data structures. For instance, finding all possible paths through a file system directory structure or a decision tree.

💡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

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

⚠️ 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: August 17, 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 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

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 "Binary Tree Paths"?

For this easy-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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