What is the time complexity of Binary Search Tree to Greater Sum Tree?

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

What data structures are used to solve Binary Search Tree to Greater Sum Tree?

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

What are the key insights for solving Binary Search Tree to Greater Sum Tree?

Insight 1: The key insight is recognizing that a reverse inorder traversal of the BST allows us to visit the nodes in descending order. Insight 2: Using a running sum allows us to efficiently accumulate the values of the larger nodes as we traverse the tree in reverse inorder.

Which companies ask LeetCode 1038. Binary Search Tree to Greater Sum Tree in interviews?

This problem is frequently asked at Datadog, SAP, eBay.

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Binary Search Tree to Greater Sum Tree - LeetCode 1038 Solution

Binary Search Tree to Greater Sum Tree - Complete Solution Guide

Binary Search Tree to Greater Sum Tree is LeetCode problem 1038, 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 Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST. As a reminder, a binary search tree is a tree that satisfies these constraints: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greater than the node's key. Both the left and right subtrees must als

Detailed Explanation

The problem requires us to convert a given Binary Search Tree (BST) into a Greater Sum Tree (GST). In a GST, each node's value is the sum of all nodes greater than or equal to it in the original BST. This means for each node, we need to add the values of all nodes to its right (larger values) and its own original value to get its new value.

Solution Approach

The provided solution uses a reverse inorder traversal of the BST. We maintain a `current_sum` variable that keeps track of the sum of all nodes visited so far in the reverse inorder. During the traversal, for each node, we add its original value to the `current_sum`, update the node's value to be the new `current_sum`, and then continue the traversal to the left subtree. This process effectively converts the BST to a GST.

Step-by-Step Algorithm

Step 1: Initialize a variable `current_sum` to 0. This will store the running sum of node values encountered during the reverse inorder traversal.
Step 2: Define a recursive function `reverse_inorder(node)` that takes a node as input.
Step 3: In the `reverse_inorder` function, check if the node is null. If it is, return (base case).
Step 4: Recursively call `reverse_inorder` on the right child of the current node. This ensures that we visit the largest values first.
Step 5: Add the current node's value to the `current_sum`.
Step 6: Update the current node's value to be the `current_sum`.
Step 7: Recursively call `reverse_inorder` on the left child of the current node.
Step 8: Call the `reverse_inorder` function with the root node to start the traversal.
Step 9: Return the modified root node.

Key Insights

Insight 1: The key insight is recognizing that a reverse inorder traversal of the BST allows us to visit the nodes in descending order.
Insight 2: Using a running sum allows us to efficiently accumulate the values of the larger nodes as we traverse the tree in reverse inorder.
Insight 3: The BST property guarantees that the right subtree contains larger values, which is crucial for the reverse inorder traversal.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: Datadog, SAP, eBay.

Binary Search Tree to Greater Sum Tree | LeetCode 1038

Medium

Tree Depth-First Search Binary Search Tree 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 {
    private int current_sum = 0;

    public TreeNode bstToGst(TreeNode root) {
        reverse_inorder(root);
        return root;
    }

    private void reverse_inorder(TreeNode node) {
        if (node == null) {
            return;
        }

        reverse_inorder(node.right);

        current_sum += node.val;
        node.val = current_sum;

        reverse_inorder(node.left);
    }
}

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Problem Info

Description

Problem 1038: Binary Search Tree to Greater Sum Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Search Tree Binary Tree

Companies

DatadogSAPeBay

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1038

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 variable `current_sum` to 0. This will store the running sum of node values encountered during the reverse inorder traversal.
2
Step 2: Define a recursive function `reverse_inorder(node)` that takes a node as input.
3
Step 3: In the `reverse_inorder` function, check if the node is null. If it is, return (base case).
4
Step 4: Recursively call `reverse_inorder` on the right child of the current node. This ensures that we visit the largest values first.
5
Step 5: Add the current node's value to the `current_sum`.
6
Step 6: Update the current node's value to be the `current_sum`.
7
Step 7: Recursively call `reverse_inorder` on the left child of the current node.
8
Step 8: Call the `reverse_inorder` function with the root node to start the traversal.
9
Step 9: Return the modified root node.

💡

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 reverse inorder traversal visits each node of the tree once. The space complexity depends on the recursion depth, which is determined by the height of the tree. In the worst case, where the tree is a linked list, the recursion depth can be equal to the number of nodes (n), leading to O(n) space. In a balanced tree, the height is log n, so the space complexity would be O(log 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 Reverse Inorder Traversal

Instead of recursion, we can use a stack to perform an iterative reverse inorder traversal. This avoids the function call overhead of recursion and can potentially reduce memory usage in some environments.

Time: O(n)Space: O(h), where h is the height of the BST. In the worst case (skewed tree), this can be O(n).

Trade-offs: The iterative approach avoids the overhead of recursion but may be slightly more complex to implement and understand. It might be preferred in environments with strict stack size limitations.

✨

Key Insights

1Insight 1: The key insight is recognizing that a reverse inorder traversal of the BST allows us to visit the nodes in descending order.
2Insight 2: Using a running sum allows us to efficiently accumulate the values of the larger nodes as we traverse the tree in reverse inorder.
3Insight 3: The BST property guarantees that the right subtree contains larger values, which is crucial for the reverse inorder traversal.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is None). The code correctly handles this because the `reverse_inorder` function checks for a null node at the beginning and returns immediately.
Edge case 2: Tree with only one node. The code correctly processes this, as the `current_sum` will be 0, then updated to the node's value, and the node's value will become the `current_sum`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Performing a regular inorder traversal instead of reverse inorder. This would result in an incorrect transformation as smaller nodes would be added to larger nodes.

✗Common mistake 2: Forgetting to update the node's value after adding it to the `current_sum`. Without this update, the tree will not be correctly transformed.

✗Common mistake 3: Not handling the base case of an empty tree correctly, which could lead to errors or infinite recursion.

✅

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: Data analysis where you need to calculate cumulative sums of values in a sorted dataset represented as a BST. This could be used for calculating cumulative sales figures, where each node represents a product and its sales.

🏦Real-world scenario 2: Ranking systems where you need to determine the rank of each element based on its value. You can represent the data as a BST and transform it to a GST to efficiently calculate ranks.

💡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 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

❓

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Binary Search Tree to Greater Sum Tree"?

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?