Farthest distance of a Node from each Node of a Tree

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++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
#define maxN 100001
  
// Adjacency List to store the graph
vector<int> adj[maxN];
  
// Stores the height of each node
int height[maxN];
  
// Stores the maximum distance of a
// node from its ancestors
int dist[maxN];
  
// Function to add edge between
// two vertices
void 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 Node
void 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 ancestor
void 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 Code
int 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;
}

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Java

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// Java program to implement
// the above approach
import 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 node
static int []height = new int[maxN];
  
// Stores the maximum distance of a
// node from its ancestors
static int []dist = new int[maxN];
  
// Function to add edge between
// two vertices
static 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 Node
static 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 ancestor
static 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 Code
public 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

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C#

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// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
static readonly int maxN = 100001;
  
// Adjacency List to store the graph
static List<int> []adj = new List<int>[maxN];
  
// Stores the height of each node
static int []height = new int[maxN];
  
// Stores the maximum distance of a
// node from its ancestors
static int []dist = new int[maxN];
  
// Function to add edge between
// two vertices
static 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 Node
static 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 ancestor
static 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 Code
public 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 

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Output: 

3 2 3 3 2 3

Time Complexity: O(V+E) 
Auxiliary Space: O(N)
 

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