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Farthest distance of a Node from each Node of a Tree

  • Difficulty Level : Hard
  • Last Updated : 15 Jun, 2021
Geek Week

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 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;
}

Java




// 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

Python3




# Python3 program to implement
# the above approach
maxN = 100001
 
# Adjacency List to store the graph
adj = [[] for i in range(maxN)]
  
# Stores the height of each node
height = [0 for i in range(maxN)]
  
# Stores the maximum distance of a
# node from its ancestors
dist = [0 for i in range(maxN)]
  
# Function to add edge between
# two vertices
def 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 Node
def 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 ancestor
def 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 Code
if __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 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

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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