What is the time complexity of Path Sum III?

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

What data structures are used to solve Path Sum III?

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 Path Sum III?

Insight 1: A brute-force approach of checking every possible path would be inefficient. We need a way to optimize the path sum calculation. Insight 2: The use of prefix sums allows us to efficiently determine if a subpath exists with the desired sum. By storing the cumulative sum from the root to each node, we can quickly calculate the sum of any path between two nodes.

Which companies ask LeetCode 437. Path Sum III in interviews?

This problem is frequently asked at Flipkart, NetApp, Zepto.

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Path Sum III - LeetCode 437 Solution

Path Sum III - Complete Solution Guide

Path Sum III is LeetCode problem 437, 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 and an integer targetSum , return the number of paths where the sum of the values along the path equals targetSum . The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes). Example 1: Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8 Output: 3 Explanation: The paths that sum to 8 are shown. Example 2: Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum =

Detailed Explanation

The problem asks us to find the number of paths in a binary tree where the sum of the node values along the path equals a given `targetSum`. A path can start and end at any node, but it must go downwards (from parent to child). This means we need to consider all possible downward paths within the tree, not just those starting at the root or ending at a leaf.

Solution Approach

The provided solutions utilize a Depth-First Search (DFS) algorithm combined with the concept of prefix sums. The DFS explores each node in the tree. A HashMap (or unordered_map in C++, or HashTable in C) is used to store the counts of different prefix sums encountered along the current path from the root to the current node. For each node, we check if there's a prefix sum encountered earlier such that the difference between the current prefix sum and that earlier prefix sum equals the target sum. If so, it means there's a path ending at the current node that sums to the target. We recursively explore the left and right subtrees, updating the prefix sum count as we go, and then decrementing the count before returning (backtracking) to ensure we only count paths relevant to the current branch. The initial entry in the hash map is 0 with a count of 1, representing an empty path that starts from the root's parent node.

Step-by-Step Algorithm

Step 1: Initialize a HashMap (or unordered_map/HashTable) to store prefix sum counts. Add an initial entry of (0, 1) to represent an empty path from before the root, which makes it possible to find paths starting from the root node itself.
Step 2: Define a recursive DFS function that takes the current node, the current path sum, the target sum, and the prefix sum counts HashMap as input.
Step 3: Base case: If the current node is null, return 0 (no path found).
Step 4: Update the current path sum by adding the value of the current node.
Step 5: Check if there exists a previous path (prefix) sum such that `current_path_sum - targetSum` equals a previously seen prefix sum. If so, it means there is a path ending at the current node with a sum equal to `targetSum`. Add the count of that prefix sum to the total count.
Step 6: Increment the count of the `current_path_sum` in the HashMap.
Step 7: Recursively call the DFS function on the left and right children of the current node, accumulating the counts from each subtree.
Step 8: Before returning from the recursive call, decrement the count of the `current_path_sum` in the HashMap. This is a backtracking step that ensures that the prefix sum counts are only relevant to the current path being explored.
Step 9: Return the total count of paths found.

Key Insights

Insight 1: A brute-force approach of checking every possible path would be inefficient. We need a way to optimize the path sum calculation.
Insight 2: The use of prefix sums allows us to efficiently determine if a subpath exists with the desired sum. By storing the cumulative sum from the root to each node, we can quickly calculate the sum of any path between two nodes.
Insight 3: The path can start at any node, not just the root. Therefore, a recursive approach that explores all possible starting points and downward paths is necessary.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Flipkart, NetApp, Zepto.

Path Sum III | LeetCode 437

Medium

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.HashMap;
import java.util.Map;

class Solution {
    public int pathSum(TreeNode root, int targetSum) {
        Map<Integer, Integer> prefixSumCounts = new HashMap<>();
        prefixSumCounts.put(0, 1);

        return dfs(root, targetSum, 0, prefixSumCounts);
    }

    private int dfs(TreeNode node, int targetSum, int currentPathSum, Map<Integer, Integer> prefixSumCounts) {
        if (node == null) {
            return 0;
        }

        currentPathSum += node.val;

        int count = prefixSumCounts.getOrDefault(currentPathSum - targetSum, 0);

        prefixSumCounts.put(currentPathSum, prefixSumCounts.getOrDefault(currentPathSum, 0) + 1);

        count += dfs(node.left, targetSum, currentPathSum, prefixSumCounts);
        count += dfs(node.right, targetSum, currentPathSum, prefixSumCounts);

        prefixSumCounts.put(currentPathSum, prefixSumCounts.get(currentPathSum) - 1);

        return count;
    }
}

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Problem Info

Description

Problem 437: Path Sum III - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

FlipkartNetAppZepto

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:437

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

9 steps

1
Step 1: Initialize a HashMap (or unordered_map/HashTable) to store prefix sum counts. Add an initial entry of (0, 1) to represent an empty path from before the root, which makes it possible to find paths starting from the root node itself.
2
Step 2: Define a recursive DFS function that takes the current node, the current path sum, the target sum, and the prefix sum counts HashMap as input.
3
Step 3: Base case: If the current node is null, return 0 (no path found).
4
Step 4: Update the current path sum by adding the value of the current node.
5
Step 5: Check if there exists a previous path (prefix) sum such that `current_path_sum - targetSum` equals a previously seen prefix sum. If so, it means there is a path ending at the current node with a sum equal to `targetSum`. Add the count of that prefix sum to the total count.
6
Step 6: Increment the count of the `current_path_sum` in the HashMap.
7
Step 7: Recursively call the DFS function on the left and right children of the current node, accumulating the counts from each subtree.
8
Step 8: Before returning from the recursive call, decrement the count of the `current_path_sum` in the HashMap. This is a backtracking step that ensures that the prefix sum counts are only relevant to the current path being explored.
9
Step 9: Return the total count of paths found.

💡

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 once. The space complexity depends on the height of the tree and the maximum number of prefix sums stored in the HashMap. In the worst case (skewed tree), this could be proportional to the number of nodes.

📖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

Brute Force

For each node, consider it as the starting point of a path. Recursively explore all downward paths from that node, summing the node values along each path. If a path sums to the targetSum, increment the count. Repeat this process for every node in the tree.

Time: O(n^2) in the worst case. For each node, we potentially traverse the entire subtree below it.Space: O(h), where h is the height of the tree. This is due to the recursion stack during DFS.

Trade-offs: This approach is simple to understand but highly inefficient for larger trees. The prefix sum optimization significantly improves the performance.

✨

Key Insights

1Insight 1: A brute-force approach of checking every possible path would be inefficient. We need a way to optimize the path sum calculation.
2Insight 2: The use of prefix sums allows us to efficiently determine if a subpath exists with the desired sum. By storing the cumulative sum from the root to each node, we can quickly calculate the sum of any path between two nodes.
3Insight 3: The path can start at any node, not just the root. Therefore, a recursive approach that explores all possible starting points and downward paths is necessary.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty Tree: The code handles the case where the root is null gracefully by returning 0, as there are no paths in an empty tree.
Edge case 2: TargetSum = 0: If the targetSum is 0, the algorithm correctly identifies paths where the sum of node values is zero. The initial `prefix_sum_counts[0] = 1` handles cases where a path starting from the root itself has a sum of zero.
Edge case 3: Nodes with negative values: The algorithm works correctly with negative node values because it uses the prefix sum approach, which handles both positive and negative values effectively.
Edge case 4: Overflow: With extremely large tree depths and node values, `current_path_sum` might potentially overflow. While the constraints suggest this is unlikely, using `long long` (or similar) in the C implementation for `current_path_sum`, `key` in the HashNode, and updating the `hash` function resolves potential integer overflow issues. This is a crucial fix for production-level code.
Edge case 5: Repeated values on path that sum to the target value. Prefix sums with counts handle this correctly by adding up the count of the prefix that results in target when differenced from current path sum

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to decrement the prefix sum count after exploring a subtree. This can lead to incorrect counts as paths from different branches might be considered together.

✗Common mistake 2: Not initializing the prefix sum count for 0 to 1. This prevents paths starting from the root node itself from being counted when their sum equals the targetSum.

✗Common mistake 3: Integer overflow: Using a simple `int` for `current_path_sum` (especially in the C implementation) may lead to integer overflows with large trees, resulting in incorrect counts. Using `long long` is essential for correctness, and the hash function must be modified accordingly.

✗Common mistake 4: Assuming that the path must start from the root or end at a leaf node. The problem explicitly states that the path can start and end at any node.

✅

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 financial transactions: Imagine a tree where each node represents a financial transaction and the value represents the amount. Finding paths with a target sum could identify sequences of transactions that add up to a specific investment or expense amount.

🏦Real-world scenario 2: Network routing: In a network represented as a tree, each node represents a router and the value represents the latency of the connection. Finding paths with a target sum could help identify routes with a specific total latency.

💡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

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

⚠️ 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: 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 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 "Path Sum III"?

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?