What is the time complexity of Second Minimum Node In a Binary Tree?

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

What data structures are used to solve Second Minimum Node In a Binary Tree?

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 Second Minimum Node In a Binary Tree?

Insight 1: The root node's value is the smallest value in the entire tree due to the tree's property. This value serves as our base for finding the second minimum. Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the tree and compare each node's value with the smallest value.

Which companies ask LeetCode 671. Second Minimum Node In a Binary Tree in interviews?

This problem is frequently asked at LinkedIn.

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Second Minimum Node In a Binary Tree - LeetCode 671 Solution

Second Minimum Node In a Binary Tree - Complete Solution Guide

Second Minimum Node In a Binary Tree is LeetCode problem 671, 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 a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes. More formally, the property root.val = min(root.left.val, root.right.val) always holds. Given such a binary tree, you need to output the second minimum value in the set made of all the nodes' value in the whole tree. If no such second minimum value e

Detailed Explanation

The problem asks us to find the second smallest value in a special binary tree. This tree has a unique property: each node's value is the minimum of its two children's values (if it has children). The root node's value is therefore the smallest value in the entire tree. We need to return the second smallest unique value. If there's no second distinct minimum value (i.e., all nodes have the same value), we should return -1. The values in the tree are non-negative.

Solution Approach

The provided Python code uses a Depth-First Search (DFS) approach to traverse the tree. It initializes `second_min` to infinity and `min_val` to the root's value. The DFS function recursively explores the tree. If a node's value is greater than `min_val`, it updates `second_min` if the node's value is smaller than the current `second_min`. After traversing the entire tree, if `second_min` is still infinity, it means there's no second minimum value, and we return -1; otherwise, we return `second_min`.

Step-by-Step Algorithm

Step 1: Initialize `second_min` to infinity and `min_val` to the root's value.
Step 2: Define a recursive DFS function that takes a node as input.
Step 3: In the DFS function, if the node is null, return.
Step 4: If the node's value is greater than `min_val`, compare it with `second_min` and update `second_min` if the node's value is smaller.
Step 5: Recursively call the DFS function on the left and right children of the node.
Step 6: After the DFS traversal is complete, check if `second_min` is still infinity. If it is, return -1; otherwise, return `second_min`.

Key Insights

Insight 1: The root node's value is the smallest value in the entire tree due to the tree's property. This value serves as our base for finding the second minimum.
Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the tree and compare each node's value with the smallest value.
Insight 3: The second minimum value must be strictly greater than the smallest value. If a node's value is greater than the smallest value and smaller than our current candidate for second minimum, we update the second minimum.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: LinkedIn.

Second Minimum Node In a Binary Tree | LeetCode 671

Easy

Tree Depth-First Search 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 {
    public int findSecondMinimumValue(TreeNode root) {
        long min1 = Long.MAX_VALUE;
        long min2 = Long.MAX_VALUE;

        dfs(root, min1, min2);

        return min2 == Long.MAX_VALUE ? -1 : (int)min2;
    }


    private void dfs(TreeNode node, long min1, long min2){
        if(node == null) return;

        if(node.val < min1){
            min2 = min1;
            min1 = node.val;
        } else if (node.val < min2 && node.val > min1){
            min2 = node.val;
        }

        dfs(node.left, min1, min2);
        dfs(node.right, min1, min2);
    }
}

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Problem Info

Description

Problem 671: Second Minimum Node In a Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:671

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 `second_min` to infinity and `min_val` to the root's value.
2
Step 2: Define a recursive DFS function that takes a node as input.
3
Step 3: In the DFS function, if the node is null, return.
4
Step 4: If the node's value is greater than `min_val`, compare it with `second_min` and update `second_min` if the node's value is smaller.
5
Step 5: Recursively call the DFS function on the left and right children of the node.
6
Step 6: After the DFS traversal is complete, check if `second_min` is still infinity. If it is, return -1; otherwise, return `second_min`.

💡

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 we visit each node at most once during the DFS traversal. The space complexity is O(h) because the maximum depth of the recursive call stack is equal to the height of 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

Breadth-First Search (BFS)

An alternative approach is to use BFS instead of DFS. We can use a queue to store the nodes to visit. The logic for updating the second minimum value remains the same.

Time: O(n)Space: O(w), where w is the maximum width of the tree. In the worst case (a complete binary tree), w can be approximately n/2, resulting in O(n) space complexity. In the best case (a skewed tree), w is 1, resulting in O(1) space complexity.

Trade-offs: BFS might be slightly more memory-intensive than DFS for deep trees but might be preferred by some. DFS utilizes stack implicitly and BFS utilizes queue explicitly. The performance characteristics would be similar in this specific context.

✨

Key Insights

1Insight 1: The root node's value is the smallest value in the entire tree due to the tree's property. This value serves as our base for finding the second minimum.
2Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the tree and compare each node's value with the smallest value.
3Insight 3: The second minimum value must be strictly greater than the smallest value. If a node's value is greater than the smallest value and smaller than our current candidate for second minimum, we update the second minimum.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: The tree has only one node (the root). The algorithm handles this by initializing `second_min` to infinity. Since no other value can be larger than the root's value, the second minimum stays infinity, and -1 is returned.
Edge case 2: All nodes in the tree have the same value. In this case, the `second_min` remains at infinity throughout the DFS traversal, and -1 is returned.
Edge case 3: The tree is a skewed tree (either left-skewed or right-skewed). The DFS traversal will still correctly identify the second minimum value, but the space complexity will be O(n).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not initializing `second_min` to infinity. This can lead to incorrect results if the second minimum value is smaller than the initial value of `second_min`.

✗Common mistake 2: Not handling the case where there is no second minimum value (i.e., all nodes have the same value). This can lead to returning an incorrect value instead of -1.

✗Common mistake 3: Incorrectly comparing the node's value with the `min_val`. The node's value should be strictly greater than `min_val` to be considered as a candidate for the second minimum.

✅

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 the second highest bidder in an auction system. Similar logic can be applied to find the second highest bid after the highest bid is determined.

🏦Real-world scenario 2: Identifying the second most popular product in an e-commerce platform. You could adapt this pattern to identify and highlight other popular offerings to users

💡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 10, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

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

@CodeNinjaJun 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 "Second Minimum Node In a Binary Tree"?

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?