What is the time complexity of Minimum Distance Between BST Nodes?

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

What data structures are used to solve Minimum Distance Between BST Nodes?

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

What are the key insights for solving Minimum Distance Between BST Nodes?

The inorder traversal of a BST yields a sorted sequence of node values. The minimum difference between any two nodes in a BST will occur between two adjacent nodes in the sorted order.

Which companies ask LeetCode 783. Minimum Distance Between BST Nodes in interviews?

This problem is frequently asked at Wix.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimum Distance Between BST Nodes - LeetCode 783 Solution

Q: How to solve Minimum Distance Between BST Nodes?

The provided solutions leverage the inorder traversal property of BSTs. They perform an inorder traversal, keeping track of the previously visited node. During the traversal, the absolute difference between the current node's value and the previous node's value is calculated. The minimum of all these differences is then returned.

Minimum Distance Between BST Nodes - Complete Solution Guide

Minimum Distance Between BST Nodes is LeetCode problem 783, 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 Search Tree (BST), return the minimum difference between the values of any two different nodes in the tree . Example 1: Input: root = [4,2,6,1,3] Output: 1 Example 2: Input: root = [1,0,48,null,null,12,49] Output: 1 Constraints: The number of nodes in the tree is in the range [2, 100] . 0 <= Node.val <= 10 5 Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/

Detailed Explanation

The problem asks us to find the minimum difference between the values of any two distinct nodes in a given Binary Search Tree (BST). A BST has the property that for any node, all nodes in its left subtree have values less than the node's value, and all nodes in its right subtree have values greater than the node's value.

Solution Approach

The provided solutions leverage the inorder traversal property of BSTs. They perform an inorder traversal, keeping track of the previously visited node. During the traversal, the absolute difference between the current node's value and the previous node's value is calculated. The minimum of all these differences is then returned.

Step-by-Step Algorithm

Step 1: Initialize a variable `min_diff` (or `minDiff`) to infinity (or maximum integer value). This variable will store the minimum difference found so far.
Step 2: Initialize a variable `prev` to negative infinity (or null if using Integer object). This variable will store the value of the previously visited node during the inorder traversal.
Step 3: Perform an inorder traversal of the BST. In the Python version, an iterative inorder traversal is used with a stack. The Java and CPP versions use recursive inorder traversal.
Step 4: During the inorder traversal, for each node, calculate the difference between the current node's value and the value of the `prev` node.
Step 5: Update `min_diff` with the minimum of its current value and the calculated difference. For the first node, prev will be negative infinity (or null) so the difference is not valid. This is why there is an 'if' condition to proceed with calculating the min_diff.
Step 6: Update `prev` to the current node's value.
Step 7: After the traversal is complete, return `min_diff`.

Key Insights

The inorder traversal of a BST yields a sorted sequence of node values.
The minimum difference between any two nodes in a BST will occur between two adjacent nodes in the sorted order.
Iterative or recursive inorder traversal can be used to generate the sorted sequence.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Wix.

Minimum Distance Between BST Nodes | LeetCode 783

Easy

Tree Depth-First Search Breadth-First Search Binary Search Tree 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 {
    Integer prev = null;
    int minDiff = Integer.MAX_VALUE;

    public int minDiffInBST(TreeNode root) {
        inorder(root);
        return minDiff;
    }

    public void inorder(TreeNode root) {
        if (root == null) return;
        inorder(root.left);
        if (prev != null) {
            minDiff = Math.min(minDiff, root.val - prev);
        }
        prev = root.val;
        inorder(root.right);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 783: Minimum Distance Between BST Nodes - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Search Tree Binary Tree

Companies

Wix

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:783

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

7 steps

1
Step 1: Initialize a variable `min_diff` (or `minDiff`) to infinity (or maximum integer value). This variable will store the minimum difference found so far.
2
Step 2: Initialize a variable `prev` to negative infinity (or null if using Integer object). This variable will store the value of the previously visited node during the inorder traversal.
3
Step 3: Perform an inorder traversal of the BST. In the Python version, an iterative inorder traversal is used with a stack. The Java and CPP versions use recursive inorder traversal.
4
Step 4: During the inorder traversal, for each node, calculate the difference between the current node's value and the value of the `prev` node.
5
Step 5: Update `min_diff` with the minimum of its current value and the calculated difference. For the first node, prev will be negative infinity (or null) so the difference is not valid. This is why there is an 'if' condition to proceed with calculating the min_diff.
6
Step 6: Update `prev` to the current node's value.
7
Step 7: After the traversal is complete, return `min_diff`.

💡

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 determined by the fact that each node in the tree is visited exactly once during the inorder traversal. The space complexity is determined by the maximum depth of the recursion stack or the stack used in the iterative inorder traversal, which in the worst-case scenario (skewed tree) is equal to the number of nodes in the tree. A balanced tree will have a space complexity of 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

Store Inorder Traversal in an Array

Perform an inorder traversal and store the node values in an array. Then, iterate through the array and find the minimum difference between adjacent elements.

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

Trade-offs: This approach requires additional space to store the inorder traversal in an array. The in-place approach used in the main solution is generally preferred when memory usage is a concern. It offers similar time complexity without additional memory overhead.

✨

Key Insights

1The inorder traversal of a BST yields a sorted sequence of node values.
2The minimum difference between any two nodes in a BST will occur between two adjacent nodes in the sorted order.
3Iterative or recursive inorder traversal can be used to generate the sorted sequence.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty Tree. The problem states there will be at least two nodes, so no need to consider the case of empty tree.
Edge case 2: Skewed Tree (all nodes on one side). The algorithm still works correctly for skewed trees, although the space complexity will be O(n) in this case, due to the recursion depth or stack size.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the BST property and trying to compare arbitrary nodes. The BST property ensures that the closest values will be adjacent in the inorder traversal.

✗Common mistake 2: Not initializing the `prev` variable correctly, leading to incorrect calculations for the first node visited. Setting it to negative infinity initially ensures correct calculation of the first difference.

✅

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: Optimizing data storage. Finding the minimum difference between data points can help in data compression and efficient storage.

🏦Real-world scenario 2: Anomaly Detection. Finding the minimum distance between elements can be used to detect anomalies, if you find distances higher than a given threshold it can be an indicator.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 3, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 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.

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 "Minimum Distance Between BST Nodes"?

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?