Given a Tree, the task is to find the farthest node from each node to another node in the given tree.
Examples
Input:
Output: 2 3 3 3 4 4 4
Explanation:
Maximum Distance from Node 1 : 2 (Nodes {5, 6, 7} are at a distance 2)
Maximum Distance from Node 2 : 3 (Nodes {6, 7} are at a distance 3)
Maximum Distance from Node 3 : 3 (Nodes {5, 6, 7} are at a distance 3)
Maximum Distance from Node 4 : 3 (Node {5} is at a distance 3)
Maximum Distance from Node 5 : 4 (Nodes {6, 7} are at a distance 4)
Maximum Distance from Node 6 : 4 (Node {5} is at a distance 4)
Maximum Distance from Node 7 : 4 (Node {5} is at a distance 4)Input:
Output : 3 2 3 3 2 3
Approach:
Follow the steps below to solve the problem:
- Calculate the height of each node of the tree (Assuming the leaf nodes are at height 1) using DFS
- This gives the maximum distance from a Node to all Nodes present in its Subtree. Store these heights.
- Now, perform DFS to calculate the maximum distance of a Node from all its ancestors. Store these distances.
- For each node, print the maximum of the two distances calculated.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;#define maxN 100001// Adjacency List to store the graphvector<int> adj[maxN];// Stores the height of each nodeint height[maxN];// Stores the maximum distance of a// node from its ancestorsint dist[maxN];// Function to add edge between// two verticesvoid addEdge(int u, int v){ // Insert edge from u to v adj[u].push_back(v); // Insert edge from v to u adj[v].push_back(u);}// Function to calculate height of// each Nodevoid dfs1(int cur, int par){ // Iterate in the adjacency // list of the current node for (auto u : adj[cur]) { if (u != par) { // Dfs for child node dfs1(u, cur); // Calculate height of nodes height[cur] = max(height[cur], height[u]); } } // Increase height height[cur] += 1;}// Function to calculate the maximum// distance of a node from its ancestorvoid dfs2(int cur, int par){ int max1 = 0; int max2 = 0; // Iterate in the adjacency // list of the current node for (auto u : adj[cur]) { if (u != par) { // Find two children // with maximum heights if (height[u] >= max1) { max2 = max1; max1 = height[u]; } else if (height[u] > max2) { max2 = height[u]; } } } int sum = 0; for (auto u : adj[cur]) { if (u != par) { // Calculate the maximum distance // with ancestor for every node sum = ((max1 == height[u]) ? max2 : max1); if (max1 == height[u]) dist[u] = 1 + max(1 + max2, dist[cur]); else dist[u] = 1 + max(1 + max1, dist[cur]); // Calculating for children dfs2(u, cur); } }}// Driver Codeint main(){ int n = 6; addEdge(1, 2); addEdge(2, 3); addEdge(2, 4); addEdge(2, 5); addEdge(5, 6); // Calculate height of // nodes of the tree dfs1(1, 0); // Calculate the maximum // distance with ancestors dfs2(1, 0); // Print the maximum of the two // distances from each node for (int i = 1; i <= n; i++) cout << (max(dist[i], height[i]) - 1) << " "; return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{static final int maxN = 100001;// Adjacency List to store the graph@SuppressWarnings("unchecked")static Vector<Integer> []adj = new Vector[maxN];// Stores the height of each nodestatic int []height = new int[maxN];// Stores the maximum distance of a// node from its ancestorsstatic int []dist = new int[maxN];// Function to add edge between// two verticesstatic void addEdge(int u, int v){ // Insert edge from u to v adj[u].add(v); // Insert edge from v to u adj[v].add(u);}// Function to calculate height of// each Nodestatic void dfs1(int cur, int par){ // Iterate in the adjacency // list of the current node for(int u : adj[cur]) { if (u != par) { // Dfs for child node dfs1(u, cur); // Calculate height of nodes height[cur] = Math.max(height[cur], height[u]); } } // Increase height height[cur] += 1;}// Function to calculate the maximum// distance of a node from its ancestorstatic void dfs2(int cur, int par){ int max1 = 0; int max2 = 0; // Iterate in the adjacency // list of the current node for(int u : adj[cur]) { if (u != par) { // Find two children // with maximum heights if (height[u] >= max1) { max2 = max1; max1 = height[u]; } else if (height[u] > max2) { max2 = height[u]; } } } int sum = 0; for(int u : adj[cur]) { if (u != par) { // Calculate the maximum distance // with ancestor for every node sum = ((max1 == height[u]) ? max2 : max1); if (max1 == height[u]) dist[u] = 1 + Math.max(1 + max2, dist[cur]); else dist[u] = 1 + Math.max(1 + max1, dist[cur]); // Calculating for children dfs2(u, cur); } }}// Driver Codepublic static void main(String[] args){ int n = 6; for(int i = 0; i < adj.length; i++) adj[i] = new Vector<Integer>(); addEdge(1, 2); addEdge(2, 3); addEdge(2, 4); addEdge(2, 5); addEdge(5, 6); // Calculate height of // nodes of the tree dfs1(1, 0); // Calculate the maximum // distance with ancestors dfs2(1, 0); // Print the maximum of the two // distances from each node for(int i = 1; i <= n; i++) System.out.print((Math.max(dist[i], height[i]) - 1) + " ");}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to implement# the above approachmaxN = 100001# Adjacency List to store the graphadj = [[] for i in range(maxN)] # Stores the height of each nodeheight = [0 for i in range(maxN)] # Stores the maximum distance of a# node from its ancestorsdist = [0 for i in range(maxN)] # Function to add edge between# two verticesdef addEdge(u, v): # Insert edge from u to v adj[u].append(v) # Insert edge from v to u adj[v].append(u)# Function to calculate height of# each Nodedef dfs1(cur, par): # Iterate in the adjacency # list of the current node for u in adj[cur]: if (u != par): # Dfs for child node dfs1(u, cur) # Calculate height of nodes height[cur] = max(height[cur], height[u]) # Increase height height[cur] += 1# Function to calculate the maximum# distance of a node from its ancestordef dfs2(cur, par): max1 = 0 max2 = 0 # Iterate in the adjacency # list of the current node for u in adj[cur]: if (u != par): # Find two children # with maximum heights if (height[u] >= max1): max2 = max1 max1 = height[u] elif (height[u] > max2): max2 = height[u] sum = 0 for u in adj[cur]: if (u != par): # Calculate the maximum distance # with ancestor for every node sum = (max2 if (max1 == height[u]) else max1) if (max1 == height[u]): dist[u] = 1 + max(1 + max2, dist[cur]) else: dist[u] = 1 + max(1 + max1, dist[cur]) # Calculating for children dfs2(u, cur) # Driver Codeif __name__=="__main__": n = 6 addEdge(1, 2) addEdge(2, 3) addEdge(2, 4) addEdge(2, 5) addEdge(5, 6) # Calculate height of # nodes of the tree dfs1(1, 0) # Calculate the maximum # distance with ancestors dfs2(1, 0) # Print the maximum of the two # distances from each node for i in range(1, n + 1): print(max(dist[i], height[i]) - 1, end = ' ') # This code is contributed by rutvik_56 |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic;class GFG{static readonly int maxN = 100001;// Adjacency List to store the graphstatic List<int> []adj = new List<int>[maxN];// Stores the height of each nodestatic int []height = new int[maxN];// Stores the maximum distance of a// node from its ancestorsstatic int []dist = new int[maxN];// Function to add edge between// two verticesstatic void addEdge(int u, int v){ // Insert edge from u to v adj[u].Add(v); // Insert edge from v to u adj[v].Add(u);}// Function to calculate height of// each Nodestatic void dfs1(int cur, int par){ // Iterate in the adjacency // list of the current node foreach(int u in adj[cur]) { if (u != par) { // Dfs for child node dfs1(u, cur); // Calculate height of nodes height[cur] = Math.Max(height[cur], height[u]); } } // Increase height height[cur] += 1;}// Function to calculate the maximum// distance of a node from its ancestorstatic void dfs2(int cur, int par){ int max1 = 0; int max2 = 0; // Iterate in the adjacency // list of the current node foreach(int u in adj[cur]) { if (u != par) { // Find two children // with maximum heights if (height[u] >= max1) { max2 = max1; max1 = height[u]; } else if (height[u] > max2) { max2 = height[u]; } } } int sum = 0; foreach(int u in adj[cur]) { if (u != par) { // Calculate the maximum distance // with ancestor for every node sum = ((max1 == height[u]) ? max2 : max1); if (max1 == height[u]) dist[u] = 1 + Math.Max(1 + max2, dist[cur]); else dist[u] = 1 + Math.Max(1 + max1, dist[cur]); // Calculating for children dfs2(u, cur); } }}// Driver Codepublic static void Main(String[] args){ int n = 6; for(int i = 0; i < adj.Length; i++) adj[i] = new List<int>(); addEdge(1, 2); addEdge(2, 3); addEdge(2, 4); addEdge(2, 5); addEdge(5, 6); // Calculate height of // nodes of the tree dfs1(1, 0); // Calculate the maximum // distance with ancestors dfs2(1, 0); // Print the maximum of the two // distances from each node for(int i = 1; i <= n; i++) Console.Write((Math.Max(dist[i], height[i]) - 1) + " ");}}// This code is contributed by Rohit_ranjan |
Javascript
<script> // JavaScript program to implement the above approach let maxN = 100001; // Adjacency List to store the graph let adj = new Array(maxN); adj.fill(0); // Stores the height of each node let height = new Array(maxN); height.fill(0); // Stores the maximum distance of a // node from its ancestors let dist = new Array(maxN); dist.fill(0); // Function to add edge between // two vertices function addEdge(u, v) { // Insert edge from u to v adj[u].push(v); // Insert edge from v to u adj[v].push(u); } // Function to calculate height of // each Node function dfs1(cur, par) { // Iterate in the adjacency // list of the current node for(let u = 0; u < adj[cur].length; u++) { if (adj[cur][u] != par) { // Dfs for child node dfs1(adj[cur][u], cur); // Calculate height of nodes height[cur] = Math.max(height[cur], height[adj[cur][u]]); } } // Increase height height[cur] += 1; } // Function to calculate the maximum // distance of a node from its ancestor function dfs2(cur, par) { let max1 = 0; let max2 = 0; // Iterate in the adjacency // list of the current node for(let u = 0; u < adj[cur].length; u++) { if (adj[cur][u] != par) { // Find two children // with maximum heights if (height[adj[cur][u]] >= max1) { max2 = max1; max1 = height[adj[cur][u]]; } else if (height[adj[cur][u]] > max2) { max2 = height[adj[cur][u]]; } } } let sum = 0; for(let u = 0; u < adj[cur].length; u++) { if (adj[cur][u] != par) { // Calculate the maximum distance // with ancestor for every node sum = ((max1 == height[adj[cur][u]]) ? max2 : max1); if (max1 == height[adj[cur][u]]) dist[adj[cur][u]] = 1 + Math.max(1 + max2, dist[cur]); else dist[adj[cur][u]] = 1 + Math.max(1 + max1, dist[cur]); // Calculating for children dfs2(adj[cur][u], cur); } } } let n = 6; for(let i = 0; i < adj.length; i++) adj[i] = []; addEdge(1, 2); addEdge(2, 3); addEdge(2, 4); addEdge(2, 5); addEdge(5, 6); // Calculate height of // nodes of the tree dfs1(1, 0); // Calculate the maximum // distance with ancestors dfs2(1, 0); // Print the maximum of the two // distances from each node for(let i = 1; i <= n; i++) document.write((Math.max(dist[i], height[i]) - 1) + " ");</script> |
Output:
3 2 3 3 2 3
Time Complexity: O(V+E)
Auxiliary Space: O(N)
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