What is the time complexity of Trim a Binary Search Tree?

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

What data structures are used to solve Trim a Binary Search 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 Trim a Binary Search Tree?

Insight 1: The BST property is crucial for efficient trimming. Nodes smaller than 'low' can be completely discarded along with their left subtrees, and nodes larger than 'high' can be discarded along with their right subtrees. Insight 2: Recursion provides a natural way to traverse the tree and modify it in place, adjusting the left and right children of each node based on the 'low' and 'high' values.

Which companies ask LeetCode 669. Trim a Binary Search Tree in interviews?

This problem is frequently asked at josh technology.

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Trim a Binary Search Tree - LeetCode 669 Solution

Trim a Binary Search Tree - Complete Solution Guide

Trim a Binary Search Tree is LeetCode problem 669, 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 and the lowest and highest boundaries as low and high , trim the tree so that all its elements lies in [low, high] . Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer . Return the root of the trimmed binary search tree . Note that the root may change depending on the given bounds. Example 1: Input: ro

Detailed Explanation

The problem asks us to trim a given Binary Search Tree (BST) such that all node values fall within a specified range [low, high]. The trimming process should preserve the relative structure of the remaining nodes. The function needs to return the root of the newly trimmed BST, which might be different from the original root if the original root's value was outside the [low, high] range.

Solution Approach

The solution uses a recursive approach to traverse the BST. For each node, it checks if the node's value is within the range [low, high]. If the value is outside the range, the function recursively calls itself on either the left or right subtree, depending on whether the value is greater than 'high' or less than 'low', respectively. If the value is within the range, the function recursively trims the left and right subtrees and then returns the current node.

Step-by-Step Algorithm

Step 1: Base Case: If the current node (root) is null, return null.
Step 2: If the current node's value is greater than 'high', the node and its right subtree are outside the range. Recursively call trimBST on the left subtree (root.left) with the same 'low' and 'high' values and return the result. This effectively removes the current node and replaces it with the trimmed left subtree.
Step 3: If the current node's value is less than 'low', the node and its left subtree are outside the range. Recursively call trimBST on the right subtree (root.right) with the same 'low' and 'high' values and return the result. This effectively removes the current node and replaces it with the trimmed right subtree.
Step 4: If the current node's value is within the range [low, high], recursively trim the left subtree by calling trimBST on root.left, and assign the result back to root.left. Similarly, recursively trim the right subtree by calling trimBST on root.right, and assign the result back to root.right.
Step 5: Return the current node (root).

Key Insights

Insight 1: The BST property is crucial for efficient trimming. Nodes smaller than 'low' can be completely discarded along with their left subtrees, and nodes larger than 'high' can be discarded along with their right subtrees.
Insight 2: Recursion provides a natural way to traverse the tree and modify it in place, adjusting the left and right children of each node based on the 'low' and 'high' values.
Insight 3: The base case for the recursion is when the current node is null; in this case, we simply return null.

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: josh technology.

Trim a Binary Search Tree | LeetCode 669

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 {
    public TreeNode trimBST(TreeNode root, int low, int high) {
        if (root == null) {
            return null;
        }

        if (root.val > high) {
            return trimBST(root.left, low, high);
        }

        if (root.val < low) {
            return trimBST(root.right, low, high);
        }

        root.left = trimBST(root.left, low, high);
        root.right = trimBST(root.right, low, high);

        return root;
    }
}

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Problem Info

Description

Problem 669: Trim a Binary Search Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Search Tree Binary Tree

Companies

josh technology

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:669

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

5 steps

1
Step 1: Base Case: If the current node (root) is null, return null.
2
Step 2: If the current node's value is greater than 'high', the node and its right subtree are outside the range. Recursively call trimBST on the left subtree (root.left) with the same 'low' and 'high' values and return the result. This effectively removes the current node and replaces it with the trimmed left subtree.
3
Step 3: If the current node's value is less than 'low', the node and its left subtree are outside the range. Recursively call trimBST on the right subtree (root.right) with the same 'low' and 'high' values and return the result. This effectively removes the current node and replaces it with the trimmed right subtree.
4
Step 4: If the current node's value is within the range [low, high], recursively trim the left subtree by calling trimBST on root.left, and assign the result back to root.left. Similarly, recursively trim the right subtree by calling trimBST on root.right, and assign the result back to root.right.
5
Step 5: Return the current node (root).

💡

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)

💾

Space Complexity

O(H)

📝

Why?

The time complexity is O(N) because, in the worst-case scenario, every node might have to be visited once. The space complexity stems from the recursive calls. Each call adds a frame to the call stack. The maximum depth of the call stack is determined by 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

Iterative Approach

An iterative approach could be used instead of recursion, using a stack to simulate the recursive calls. This would avoid the call stack overhead.

Time: O(N)Space: O(H), where H is the height of the tree, due to the stack.

Trade-offs: The iterative approach might be slightly more complex to implement but could be preferred if stack overflow is a concern for very deep trees.

✨

Key Insights

1Insight 1: The BST property is crucial for efficient trimming. Nodes smaller than 'low' can be completely discarded along with their left subtrees, and nodes larger than 'high' can be discarded along with their right subtrees.
2Insight 2: Recursion provides a natural way to traverse the tree and modify it in place, adjusting the left and right children of each node based on the 'low' and 'high' values.
3Insight 3: The base case for the recursion is when the current node is null; in this case, we simply return null.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is null). The code handles this by returning null in the base case.
Edge case 2: All nodes are outside the range. The code will recursively trim the tree, eventually returning null, indicating an empty tree.
Edge case 3: All nodes are within the range. The code will recursively trim the left and right subtrees, leaving the tree structure intact.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the base case (root == null), which would lead to a null pointer exception or infinite recursion.

✗Common mistake 2: Not updating the root's left or right children after trimming. If the recursive call returns a different node, the link must be updated (e.g., `root.left = trimBST(root.left, low, high)`).

✗Common mistake 3: Incorrectly implementing the trimming logic, e.g., not using the BST property to determine which subtree to trim.

✅

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 filtering in a database. Imagine a database storing numerical values with indexes implemented as BSTs. This algorithm can be adapted to efficiently filter data within a specific range.

🏦Real-world scenario 2: Range queries in search engines. BSTs can be used to index search terms. Trimming can be useful in narrowing down search results to a specific range of values (e.g., price range, date range).

💡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 8, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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 "Trim a Binary Search 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.

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