What is the time complexity of Find Mode in Binary Search Tree?

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

What data structures are used to solve Find Mode in Binary Search Tree?

This problem uses Tree, Depth-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 Find Mode in Binary Search Tree?

Insight 1: Inorder traversal of a BST yields a sorted sequence of node values. Insight 2: Keeping track of the current value, its count, and the maximum count encountered so far allows us to determine the modes efficiently during traversal.

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Find Mode in Binary Search Tree - LeetCode 501 Solution

Find Mode in Binary Search Tree - Complete Solution Guide

Find Mode in Binary Search Tree is LeetCode problem 501, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Framing

Find Mode in Binary Search Tree is a Easy LeetCode problem that rewards careful tracing, edge-case handling, and a clear grasp of Tree and Depth-First Search. The best solutions usually explain why the chosen invariant holds before they optimize for time or space.

Quick Example Mindset

A useful way to test Find Mode in Binary Search Tree is to start with a tiny input that exposes the boundary conditions, then run the same logic on a slightly larger case to verify the tree behavior and the depth-first search interaction. That second pass is where off-by-one mistakes and missing updates usually appear.

Problem Statement

Given the root of a binary search tree (BST) with duplicates, return all the mode(s) (i.e., the most frequently occurred element) in it . If the tree has more than one mode, return them in any order . Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than or equal to the node's key. The right subtree of a node contains only nodes with keys greater than or equal to the node's key. Both the left and right subtrees must also be binary search trees. Ex

Detailed Explanation

The problem asks us to find the mode(s) of a Binary Search Tree (BST). The mode is the value that appears most frequently in the tree. Since the tree is a BST, we can leverage the inorder traversal to visit the nodes in sorted order, which will help us efficiently count the frequency of each value. The output should be a list of all modes. It's important to note that there can be multiple modes in a tree if multiple values have the same maximum frequency.

Solution Approach

The solution uses an inorder traversal of the BST to visit nodes in sorted order. It maintains a count of the current value being visited. If the current value changes, the count is reset to 1. The maximum count seen so far is also tracked. Whenever the current count exceeds the maximum count, the result list is cleared, and the current value is added as the new mode. If the current count equals the maximum count, the current value is added to the existing modes.

Step-by-Step Algorithm

Step 1: Initialize variables: `count` dictionary (Python/C++) or `maxCount`, `currCount`, `prevVal`, `modes` (Java/C) to track frequency and modes.
Step 2: Perform an inorder traversal of the BST.
Step 3: During traversal, compare the current node's value with the previous value (or use a dictionary to store counts).
Step 4: If the current value is the same as the previous value, increment the current count.
Step 5: If the current value is different from the previous value, reset the current count to 1.
Step 6: Update the maximum count seen so far. If current count exceeds the maxCount, reset the `modes` list and add the current value. If current count equals maxCount, add the current value to the `modes` list.
Step 7: After the traversal is complete, return the list of modes.

Key Insights

Insight 1: Inorder traversal of a BST yields a sorted sequence of node values.
Insight 2: Keeping track of the current value, its count, and the maximum count encountered so far allows us to determine the modes efficiently during traversal.
Insight 3: We need to handle the edge case where multiple values have the same maximum frequency, adding all of them to the result.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Study Paths

Continue from this problem into the surrounding topic and company clusters to compare how the same pattern appears in other interview settings.

Related topics: Tree, Depth-First Search, Binary Search Tree, Binary Tree

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.

Find Mode in Binary Search Tree | LeetCode 501

Easy

Tree Depth-First Search Binary Search Tree Binary Tree

Study Path

Browse topics

Continue through the pattern hub for deeper coverage and neighboring problem variants.

Tree Depth-First Search Binary Search Tree Binary Tree

Company Signals

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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 {
    int maxCount = 0;
    int currCount = 0;
    int prevVal = Integer.MIN_VALUE;
    List<Integer> modes = new ArrayList<>();

    public int[] findMode(TreeNode root) {
        inorder(root);
        int[] result = new int[modes.size()];
        for (int i = 0; i < modes.size(); i++) {
            result[i] = modes.get(i);
        }
        return result;
    }

    private void inorder(TreeNode node) {
        if (node == null) return;
        inorder(node.left);
        if (node.val == prevVal) {
            currCount++;
        } else {
            currCount = 1;
            prevVal = node.val;
        }
        if (currCount > maxCount) {
            maxCount = currCount;
            modes.clear();
            modes.add(node.val);
        } else if (currCount == maxCount) {
            modes.add(node.val);
        }
        inorder(node.right);
    }
}

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Problem Info

Description

Problem 501: Find Mode in Binary Search Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Search Tree Binary Tree

Stats

Difficulty:Easy

Languages:4

Problem #:501

Problem Explanation

🧭

Next Steps

Use this problem as a starting point, then move into adjacent patterns and company sets to reinforce the same idea in different contexts.

Topic hub

Study neighboring problems by pattern and difficulty.

Company hub

See where this problem appears in interview loops.

Pattern guides

Read deeper walkthroughs of the algorithm family.

💡

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 variables: `count` dictionary (Python/C++) or `maxCount`, `currCount`, `prevVal`, `modes` (Java/C) to track frequency and modes.
2
Step 2: Perform an inorder traversal of the BST.
3
Step 3: During traversal, compare the current node's value with the previous value (or use a dictionary to store counts).
4
Step 4: If the current value is the same as the previous value, increment the current count.
5
Step 5: If the current value is different from the previous value, reset the current count to 1.
6
Step 6: Update the maximum count seen so far. If current count exceeds the maxCount, reset the `modes` list and add the current value. If current count equals maxCount, add the current value to the `modes` list.
7
Step 7: After the traversal is complete, return the list of modes.

💡

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 O(n) because the inorder traversal visits each node once. The space complexity is O(n) primarily due to the hash map or ArrayList that stores counts or the list of modes, which can potentially store all unique node values if they all appear with the same frequency. The recursive stack used by inorder traversal can reach O(h) height, which can also be O(n) in the worst case (a skewed 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

Using a HashMap/Dictionary for Counting

This approach involves traversing the tree and using a HashMap (or dictionary in Python) to store the frequency of each node's value. After the traversal, iterate through the HashMap to find the mode(s). This is what the python and cpp solutions are doing.

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

Trade-offs: Simple to implement but uses extra space to store the frequency of each value. This approach is more suitable when the number of unique values is relatively small compared to the total number of nodes in the tree.

Morris Traversal (Threaded Binary Tree)

Morris traversal can be used to traverse the tree in O(1) space. However, it makes the problem more complex to solve because the inorder traversal is no longer straightforward.

Time: O(n)Space: O(1)

Trade-offs: Reduces space complexity to O(1), but significantly increases the complexity of the implementation. This approach might be useful if memory is extremely limited.

✨

Key Insights

1Insight 1: Inorder traversal of a BST yields a sorted sequence of node values.
2Insight 2: Keeping track of the current value, its count, and the maximum count encountered so far allows us to determine the modes efficiently during traversal.
3Insight 3: We need to handle the edge case where multiple values have the same maximum frequency, adding all of them to the result.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree. The solutions correctly return an empty list in this case.
Edge case 2: Tree with only one node. The solution correctly returns a list containing only that node's value.
Edge case 3: Tree where all nodes have the same value. The solutions correctly identify that value as the mode.
Edge case 4: Tree with multiple modes. The solutions correctly identify all values that have the same maximum frequency.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to clear the result list when a new maximum count is encountered. This would lead to incorrect modes being returned.

✗Common mistake 2: Not handling the case where multiple values have the same maximum frequency. The solution needs to identify *all* such values and include them in the result.

✗Common mistake 3: Not accounting for the skewed tree (worst-case height equals number of nodes) when considering recursion depth as space complexity.

✅

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 website traffic to find the most frequently visited pages.

🏦Real-world scenario 2: Identifying the most popular product categories in an e-commerce platform.

🎮Real-world scenario 3: Analyzing sensor data to determine the most common readings.

💡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: October 20, 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 21

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

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

What should I learn from easy problems?

What is the importance of time and space complexity analysis?

What is the optimal approach for "Find Mode in Binary Search 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.

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