What is the time complexity of Create Binary Tree From Descriptions?

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

What data structures are used to solve Create Binary Tree From Descriptions?

This problem uses Array, Hash Table, Tree, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Create Binary Tree From Descriptions?

Insight 1: The core idea is to use a hash map (or dictionary) to store nodes and their corresponding TreeNode objects. This allows efficient access and manipulation of nodes based on their values. Insight 2: Identifying the root node is crucial. The root is the only node that is not a child of any other node. This can be determined by tracking all child nodes and finding the node that isn't in the set of children.

Which companies ask LeetCode 2196. Create Binary Tree From Descriptions in interviews?

This problem is frequently asked at Clari, LinkedIn.

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Create Binary Tree From Descriptions - LeetCode 2196 Solution

Create Binary Tree From Descriptions - Complete Solution Guide

Create Binary Tree From Descriptions is LeetCode problem 2196, 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

You are given a 2D integer array descriptions where descriptions[i] = [parent i , child i , isLeft i ] indicates that parent i is the parent of child i in a binary tree of unique values. Furthermore, If isLeft i == 1 , then child i is the left child of parent i . If isLeft i == 0 , then child i is the right child of parent i . Construct the binary tree described by descriptions and return its root . The test cases will be generated such that the binary tree is valid . Example 1: Input: descripti

Detailed Explanation

The problem requires constructing a binary tree from a given array of descriptions. Each description provides information about a parent-child relationship, including whether the child is the left or right child of the parent. The input is a 2D integer array `descriptions`, where each row represents a parent-child relationship: `[parent_i, child_i, isLeft_i]`. The goal is to return the root node of the constructed binary tree. The tree contains unique values, and the provided descriptions guarantee a valid binary tree.

Solution Approach

The solution uses a hash map (`nodes_map`) to store `TreeNode` objects, keyed by their values. It also uses a set (`children_set`) to track all the child nodes. The algorithm iterates through the `descriptions` array. For each description, it creates `TreeNode` objects for the parent and child (if they don't already exist in the `nodes_map`), adds the child to `children_set`, and establishes the parent-child relationship (left or right) based on `isLeft`. After processing all descriptions, the algorithm iterates through the descriptions again to find the root node. It identifies the root as the node whose value is not present in the `children_set`. Finally, it returns the `TreeNode` object corresponding to the root value from `nodes_map`.

Step-by-Step Algorithm

Step 1: Initialize a hash map `nodes_map` to store the tree nodes (TreeNode objects) with the node values as keys. Initialize a set `children_set` to store the values of all child nodes.
Step 2: Iterate through the `descriptions` array.
Step 3: For each description `[parent_val, child_val, is_left]`, add `child_val` to the `children_set`.
Step 4: If `parent_val` is not in `nodes_map`, create a new `TreeNode` with value `parent_val` and add it to `nodes_map`.
Step 5: If `child_val` is not in `nodes_map`, create a new `TreeNode` with value `child_val` and add it to `nodes_map`.
Step 6: Set the left or right child of the parent node in `nodes_map` to the child node, based on the value of `is_left` (1 for left, 0 for right).
Step 7: After processing all descriptions, iterate through the `descriptions` array again.
Step 8: For each description, check if the `parent_val` is present in the `children_set`. If not, it means `parent_val` is the root node's value.
Step 9: Return the `TreeNode` object from `nodes_map` that corresponds to the root node's value.

Key Insights

Insight 1: The core idea is to use a hash map (or dictionary) to store nodes and their corresponding TreeNode objects. This allows efficient access and manipulation of nodes based on their values.
Insight 2: Identifying the root node is crucial. The root is the only node that is not a child of any other node. This can be determined by tracking all child nodes and finding the node that isn't in the set of children.
Insight 3: The constraint that the tree is valid simplifies the process because we do not have to handle invalid inputs such as cycles or multiple roots.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Tree, Binary Tree.

Companies

Asked at: Clari, LinkedIn.

Create Binary Tree From Descriptions | LeetCode 2196

Medium

Array Hash Table 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;
 *     }
 * }
 */
import java.util.*;

class Solution {
    public TreeNode createBinaryTree(int[][] descriptions) {
        Map<Integer, TreeNode> nodes_map = new HashMap<>();
        Set<Integer> children_set = new HashSet<>();

        for (int[] description : descriptions) {
            int parent_val = description[0];
            int child_val = description[1];
            int is_left = description[2];

            children_set.add(child_val);

            if (!nodes_map.containsKey(parent_val)) {
                nodes_map.put(parent_val, new TreeNode(parent_val));
            }

            if (!nodes_map.containsKey(child_val)) {
                nodes_map.put(child_val, new TreeNode(child_val));
            }

            if (is_left == 1) {
                nodes_map.get(parent_val).left = nodes_map.get(child_val);
            } else {
                nodes_map.get(parent_val).right = nodes_map.get(child_val);
            }
        }

        int root_val = -1;
        for (int[] description : descriptions) {
            int parent_val = description[0];
            if (!children_set.contains(parent_val)) {
                root_val = parent_val;
                break;
            }
        }

        return nodes_map.get(root_val);
    }
}

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Problem Info

Description

Problem 2196: Create Binary Tree From Descriptions - Detailed explanation and solution approaches.

Topics

Array Hash Table Tree Binary Tree

Companies

ClariLinkedIn

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2196

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 hash map `nodes_map` to store the tree nodes (TreeNode objects) with the node values as keys. Initialize a set `children_set` to store the values of all child nodes.
2
Step 2: Iterate through the `descriptions` array.
3
Step 3: For each description `[parent_val, child_val, is_left]`, add `child_val` to the `children_set`.
4
Step 4: If `parent_val` is not in `nodes_map`, create a new `TreeNode` with value `parent_val` and add it to `nodes_map`.
5
Step 5: If `child_val` is not in `nodes_map`, create a new `TreeNode` with value `child_val` and add it to `nodes_map`.
6
Step 6: Set the left or right child of the parent node in `nodes_map` to the child node, based on the value of `is_left` (1 for left, 0 for right).
7
Step 7: After processing all descriptions, iterate through the `descriptions` array again.
8
Step 8: For each description, check if the `parent_val` is present in the `children_set`. If not, it means `parent_val` is the root node's value.
9
Step 9: Return the `TreeNode` object from `nodes_map` that corresponds to the root node's value.

💡

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 two iterations over the `descriptions` array. The hash map operations are assumed to be constant time on average. The space complexity is proportional to the number of unique nodes in the tree, which can be at most equal to the number of descriptions. The children set can store up to n distinct children.

📖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 parent map

Instead of tracking children, maintain a map of child to parent. The root is the node that is not a key in the child-to-parent map.

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

Trade-offs: This alternative is very similar in performance but might be slightly easier to understand for some people. Both are effectively the same in terms of overall efficiency.

✨

Key Insights

1Insight 1: The core idea is to use a hash map (or dictionary) to store nodes and their corresponding TreeNode objects. This allows efficient access and manipulation of nodes based on their values.
2Insight 2: Identifying the root node is crucial. The root is the only node that is not a child of any other node. This can be determined by tracking all child nodes and finding the node that isn't in the set of children.
3Insight 3: The constraint that the tree is valid simplifies the process because we do not have to handle invalid inputs such as cycles or multiple roots.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: The input descriptions might be empty. In this case, the code should correctly return `None` (or `null` in Java/C++). The given code assumes there's at least one description.
Edge case 2: The root node has a single child. The solution handles this correctly by iterating through the descriptions to build the binary tree and identify the root.
Edge case 3: All nodes are arranged in a single chain (e.g., a linked list implemented as a binary tree). The algorithm correctly processes this structure and identifies the root.
Edge case 4: Large input size (close to the constraint limits). The hash map ensures reasonably efficient lookup and insertion, even with a large number of nodes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly identifying the root node. Beginners might assume the first parent in the `descriptions` array is the root, which is incorrect.

✗Common mistake 2: Forgetting to create `TreeNode` objects for both the parent and child nodes if they don't already exist in the hash map. This leads to incorrect tree construction.

✗Common mistake 3: Confusing left and right child assignments. Carefully check the `isLeft` flag when establishing parent-child relationships.

✗Common mistake 4: Not handling the edge case where the `descriptions` array is empty. The current solutions assume that descriptions is never empty, but in a real-world scenario, checking for `descriptions` being empty and returning `None`/`null` is necessary.

✅

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: Representing hierarchical data structures, such as organizational charts or file systems. The `descriptions` array could represent relationships between employees or files and directories.

🏦Real-world scenario 2: Parsing and building abstract syntax trees (ASTs) in compilers. The `descriptions` array could represent the grammar rules used to build the AST from source code.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: July 7, 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 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 "Create Binary Tree From Descriptions"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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