What is the time complexity of All Nodes Distance K in Binary Tree?

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

What data structures are used to solve All Nodes Distance K in Binary Tree?

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

What are the key insights for solving All Nodes Distance K in Binary Tree?

Insight 1: The problem requires calculating distances in a tree, which is not directly supported by the tree structure itself. Converting the tree into a graph allows traversal in both parent-child and child-parent directions. Insight 2: Breadth-First Search (BFS) is an effective way to traverse the graph and find all nodes at a specific distance from the target node.

Which companies ask LeetCode 863. All Nodes Distance K in Binary Tree in interviews?

This problem is frequently asked at DE Shaw, DP world, Flipkart, Nutanix, Samsung and 2 more companies.

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All Nodes Distance K in Binary Tree - LeetCode 863 Solution

All Nodes Distance K in Binary Tree - Complete Solution Guide

All Nodes Distance K in Binary Tree is LeetCode problem 863, 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 tree, the value of a target node target , and an integer k , return an array of the values of all nodes that have a distance k from the target node. You can return the answer in any order . Example 1: Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, k = 2 Output: [7,4,1] Explanation: The nodes that are a distance 2 from the target node (with value 5) have values 7, 4, and 1. Example 2: Input: root = [1], target = 1, k = 3 Output: [] Constraints: The number of n

Detailed Explanation

The problem asks us to find all nodes in a binary tree that are exactly 'k' distance away from a given 'target' node. The input includes the root of the binary tree, the target node, and the distance 'k'. The output should be a list of node values that are at distance 'k' from the target node. The tree nodes have unique values, and the distance 'k' is a non-negative integer.

Solution Approach

The solution involves two main steps: first, transforming the binary tree into an undirected graph, and second, using Breadth-First Search (BFS) to traverse the graph starting from the target node and finding all nodes that are 'k' distance away. The graph representation allows us to move up and down the tree from any node, which is necessary since we need to find nodes in all directions.

Step-by-Step Algorithm

Step 1: Build a graph (adjacency list) representation of the binary tree. Each node in the tree becomes a node in the graph, and an edge is created between a parent and its children (and vice versa), representing bidirectional connections.
Step 2: Initialize a Breadth-First Search (BFS) queue with the target node and a distance of 0.
Step 3: Initialize a 'visited' set to keep track of visited nodes during the BFS traversal to prevent cycles.
Step 4: While the queue is not empty, dequeue a node and its distance from the queue.
Step 5: If the distance of the current node is equal to 'k', add its value to the result list.
Step 6: For each neighbor of the current node in the graph, if the neighbor has not been visited, add it to the queue with a distance incremented by 1 and mark it as visited.
Step 7: Repeat steps 4-6 until the queue is empty.
Step 8: Return the list of node values that are 'k' distance away from the target.

Key Insights

Insight 1: The problem requires calculating distances in a tree, which is not directly supported by the tree structure itself. Converting the tree into a graph allows traversal in both parent-child and child-parent directions.
Insight 2: Breadth-First Search (BFS) is an effective way to traverse the graph and find all nodes at a specific distance from the target node.
Insight 3: Visited set or hashmap is crucial to prevent infinite loops when traversing the graph formed from the tree, as nodes now have connections to both their children and their parent.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: DE Shaw, DP world, Flipkart, Nutanix, Samsung, Walmart Labs, Wix.

All Nodes Distance K in Binary Tree | LeetCode 863

Medium

Hash Table Tree Depth-First Search Breadth-First Search 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(int x) { val = x; }
 * }
 */
import java.util.*;

class Solution {
    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        if (k == 0) {
            return Arrays.asList(target.val);
        }
        
        Map<TreeNode, List<TreeNode>> graph = new HashMap<>();
        
        Queue<TreeNode> q_build = new LinkedList<>();
        q_build.offer(root);
        while (!q_build.isEmpty()) {
            TreeNode node = q_build.poll();
            if (node.left != null) {
                graph.computeIfAbsent(node, key -> new ArrayList<>()).add(node.left);
                graph.computeIfAbsent(node.left, key -> new ArrayList<>()).add(node);
                q_build.offer(node.left);
            }
            if (node.right != null) {
                graph.computeIfAbsent(node, key -> new ArrayList<>()).add(node.right);
                graph.computeIfAbsent(node.right, key -> new ArrayList<>()).add(node);
                q_build.offer(node.right);
            }
        }
        
        List<Integer> result = new ArrayList<>();
        Set<TreeNode> visited = new HashSet<>();
        visited.add(target);
        Queue<Pair<TreeNode, Integer>> q_search = new LinkedList<>();
        q_search.offer(new Pair<>(target, 0));
        
        while (!q_search.isEmpty()) {
            Pair<TreeNode, Integer> current = q_search.poll();
            TreeNode current_node = current.getKey();
            int distance = current.getValue();
            
            if (distance == k) {
                result.add(current_node.val);
                continue;
            }
            
            if (graph.containsKey(current_node)) {
                for (TreeNode neighbor : graph.get(current_node)) {
                    if (!visited.contains(neighbor)) {
                        visited.add(neighbor);
                        q_search.offer(new Pair<>(neighbor, distance + 1));
                    }
                }
            }
        }
        
        return result;
    }
    
    static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

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Problem Info

Description

Problem 863: All Nodes Distance K in Binary Tree - Detailed explanation and solution approaches.

Topics

Hash Table Tree Depth-First Search Breadth-First Search Binary Tree

Companies

DE ShawDP worldFlipkartNutanixSamsungWalmart LabsWix

Stats

Difficulty:Medium

Languages:4

Companies:7

Problem #:863

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

8 steps

1
Step 1: Build a graph (adjacency list) representation of the binary tree. Each node in the tree becomes a node in the graph, and an edge is created between a parent and its children (and vice versa), representing bidirectional connections.
2
Step 2: Initialize a Breadth-First Search (BFS) queue with the target node and a distance of 0.
3
Step 3: Initialize a 'visited' set to keep track of visited nodes during the BFS traversal to prevent cycles.
4
Step 4: While the queue is not empty, dequeue a node and its distance from the queue.
5
Step 5: If the distance of the current node is equal to 'k', add its value to the result list.
6
Step 6: For each neighbor of the current node in the graph, if the neighbor has not been visited, add it to the queue with a distance incremented by 1 and mark it as visited.
7
Step 7: Repeat steps 4-6 until the queue is empty.
8
Step 8: Return the list of node values that are 'k' distance away from the target.

💡

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 both the graph construction (BFS) and the distance search (BFS) visit each node at most a constant number of times. The space complexity is O(n) because the graph and queue could store all the nodes of the tree in the worst-case scenarios.

📖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

Depth-First Search (DFS) with parent pointers

Instead of building a graph, we can use DFS to find the distance to the target node from every other node. To do this, we first find the target node using DFS and create parent pointers for each node along the way. Then, starting from the target, we can use DFS to find all nodes at distance k. This avoids explicit graph creation.

Time: O(n)Space: O(n) in the worst case (skewed tree), due to the recursion stack and parent pointers storage.

Trade-offs: DFS with parent pointers can be slightly more space-efficient in some cases as we don't store a full graph, but it's also often more complex to implement correctly. The primary solution using a graph is generally considered more readable and maintainable.

✨

Key Insights

1Insight 1: The problem requires calculating distances in a tree, which is not directly supported by the tree structure itself. Converting the tree into a graph allows traversal in both parent-child and child-parent directions.
2Insight 2: Breadth-First Search (BFS) is an effective way to traverse the graph and find all nodes at a specific distance from the target node.
3Insight 3: Visited set or hashmap is crucial to prevent infinite loops when traversing the graph formed from the tree, as nodes now have connections to both their children and their parent.

📚

Key Terms Explained

For Students

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: k = 0. The solution directly returns the target node's value in this case.
Edge case 2: Target node is the root. The BFS correctly handles this scenario.
Edge case 3: k is larger than the height of the tree. The BFS will only visit nodes within the tree structure and will return an empty list if no nodes are found at the specified distance.
Edge case 4: Empty tree (root is None). The code will likely throw an exception if not handled. The provided examples state that the number of nodes is in the range [1,500], thus not accounting for an empty tree. In practice, the root should be checked to be not None before proceeding.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling cycles in the graph traversal. Without a 'visited' set, the BFS will enter an infinite loop when moving between parent and child nodes.

✗Common mistake 2: Incorrectly calculating the distance. It's important to increment the distance correctly when enqueuing neighbor nodes in the BFS.

✗Common mistake 3: Failing to consider nodes that are ancestors of the target node when searching for nodes at distance k.

✅

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: Social network friend recommendations. Finding users within a certain 'degree of separation' from a target user can be modeled using a graph and finding nodes at a particular distance.

🏦Real-world scenario 2: Network routing. Determining the optimal path between two devices in a network can involve finding nodes at a certain distance (hop count) from a source node.

🎮Real-world scenario 3: Geneology: Determining relatives within a certain degree of separation from an ancestor, where the family tree is modelled as a graph.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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 22, 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 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "All Nodes Distance K in Binary Tree"?

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

🤖 Need help with this problem?