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Calculate number of nodes in all subtrees | Using DFS

  • Difficulty Level : Medium
  • Last Updated : 30 Jun, 2021

Given a tree in the form of adjacency list we have to calculate the number of nodes in the subtree of each node while calculating the number of nodes in the subtree of a particular node that node will also be added as a node in subtree hence the number of nodes in subtree of leaves is 1. 
Examples: 
 

Input : Consider the Graph mentioned below:

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Output : Nodes in subtree of 1 : 5
         Nodes in subtree of 2 : 1
         Nodes in subtree of 3 : 1
         Nodes in subtree of 4 : 3
         Nodes in subtree of 5 : 1

Input : Consider the Graph mentioned below:



Output : Nodes in subtree of 1 : 7
         Nodes in subtree of 2 : 2
         Nodes in subtree of 3 : 1
         Nodes in subtree of 4 : 3
         Nodes in subtree of 5 : 1
         Nodes in subtree of 6 : 1
         Nodes in subtree of 7 : 1

Explanation: First we should calculate value count[s] : the number of nodes in subtree of node s. Where subtree contains the node itself and all the nodes in the subtree of its children. Thus, we can calculate the number of nodes recursively using the concept of DFS and DP, where we should process each edge only once and count[] value of children used in calculating count[] of its parent expressing the concept of DP(Dynamic programming). 
Time Complexity : O(n) [in processing of all (n-1) edges].
 

Algorithm :
void numberOfNodes(int s, int e)
{
    vector::iterator u;
    count1[s] = 1;
    for (u = adj[s].begin(); u != adj[s].end(); u++)
    {
        // condition to omit reverse path
        // path from children to parent 
        if (*u == e)
            continue;
        
        // recursive call for DFS
        numberOfNodes(*u, s);
        
        // update count[] value of parent using
        // its children
        count1[s] += count1[*u];
    }
}

 

 

C++




// CPP code to find number of nodes
// in subtree of each node
#include <bits/stdc++.h>
using namespace std;
 
const int N = 8;
 
// variables used to store data globally
int count1[N];
 
// adjacency list representation of tree
vector<int> adj[N];
 
// function to calculate no. of nodes in a subtree
void numberOfNodes(int s, int e)
{
    vector<int>::iterator u;
    count1[s] = 1;
    for (u = adj[s].begin(); u != adj[s].end(); u++) {
         
        // condition to omit reverse path
        // path from children to parent
        if (*u == e)
            continue;
         
        // recursive call for DFS
        numberOfNodes(*u, s);
         
        // update count[] value of parent using
        // its children
        count1[s] += count1[*u];
    }
}
 
// function to add edges in graph
void addEdge(int a, int b)
{
    adj[a].push_back(b);
    adj[b].push_back(a);
}
 
// function to print result
void printNumberOfNodes()
{
    for (int i = 1; i < N; i++) {
        cout << "\nNodes in subtree of " << i;
        cout << ": " << count1[i];
    }
}
 
// driver function
int main()
{
    // insertion of nodes in graph
    addEdge(1, 2);
    addEdge(1, 4);
    addEdge(1, 5);
    addEdge(2, 6);
    addEdge(4, 3);
    addEdge(4, 7);
     
    // call to perform dfs calculation
    // making 1  as root of tree
    numberOfNodes(1, 0);
     
    // print result
    printNumberOfNodes();
    return 0;
}

Java




// A Java code to find number of nodes
// in subtree of each node
import java.util.ArrayList;
 
public class NodesInSubtree
{
    // variables used to store data globally
    static final int N = 8;
    static int count1[] = new int[N];
     
    // adjacency list representation of tree
    static ArrayList<Integer> adj[] = new ArrayList[N];
     
    // function to calculate no. of nodes in a subtree
    static void numberOfNodes(int s, int e)
    {
        count1[s] = 1;
        for(Integer u: adj[s])
        {
            // condition to omit reverse path
            // path from children to parent
            if(u == e)
                continue;
             
            // recursive call for DFS
            numberOfNodes(u ,s);
             
            // update count[] value of parent using
            // its children
            count1[s] += count1[u];
        }
    }
     
    // function to add edges in graph
    static void addEdge(int a, int b)
    {
        adj[a].add(b);
        adj[b].add(a);
    }
     
    // function to print result
    static void printNumberOfNodes()
    {
        for (int i = 1; i < N; i++)
            System.out.println("Node of a subtree of "+ i+
                                       " : "+ count1[i]);
    }
     
    // Driver function
    public static void main(String[] args)
    {
        // Creating list for all nodes
        for(int i = 0; i < N; i++)
            adj[i] = new ArrayList<>();
             
        // insertion of nodes in graph
        addEdge(1, 2);
        addEdge(1, 4);
        addEdge(1, 5);
        addEdge(2, 6);
        addEdge(4, 3);
        addEdge(4, 7);
         
        // call to perform dfs calculation
        // making 1  as root of tree
        numberOfNodes(1, 0);
         
        // print result
        printNumberOfNodes();
             
    }
 
}
// This code is contributed by Sumit Ghosh

Python3




# Python3 code to find the number of
# nodes in the subtree of each node
N = 8
 
# variables used to store data globally
count1 = [0] * (N)
 
# Adjacency list representation of tree
adj = [[] for i in range(N)]
 
# Function to calculate no. of
# nodes in subtree
def numberOfNodes(s, e):
 
    count1[s] = 1
    for u in adj[s]:
         
        # Condition to omit reverse path
        # path from children to parent
        if u == e:
            continue
         
        # recursive call for DFS
        numberOfNodes(u, s)
         
        # update count[] value of parent
        # using its children
        count1[s] += count1[u]
 
# Function to add edges in graph
def addEdge(a, b):
 
    adj[a].append(b)
    adj[b].append(a)
 
# Function to print result
def printNumberOfNodes():
 
    for i in range(1, N):
        print("Nodes in subtree of", i,
                        ":", count1[i])
 
# Driver Code
if __name__ == "__main__":
 
    # insertion of nodes in graph
    addEdge(1, 2)
    addEdge(1, 4)
    addEdge(1, 5)
    addEdge(2, 6)
    addEdge(4, 3)
    addEdge(4, 7)
     
    # call to perform dfs calculation
    # making 1 as root of tree
    numberOfNodes(1, 0)
     
    # print result
    printNumberOfNodes()
     
# This code is contributed by Rituraj Jain

C#




// C# code to find number of nodes
// in subtree of each node
using System;
using System.Collections.Generic;
class GFG
{
    // variables used to store data globally
    static readonly int N = 8;
    static int []count1 = new int[N];
     
    // adjacency list representation of tree
    static List<int> []adj = new List<int>[N];
     
    // function to calculate no. of nodes in a subtree
    static void numberOfNodes(int s, int e)
    {
        count1[s] = 1;
        foreach(int u in adj[s])
        {
            // condition to omit reverse path
            // path from children to parent
            if(u == e)
                continue;
             
            // recursive call for DFS
            numberOfNodes(u, s);
             
            // update count[] value of parent using
            // its children
            count1[s] += count1[u];
        }
    }
     
    // function to add edges in graph
    static void addEdge(int a, int b)
    {
        adj[a].Add(b);
        adj[b].Add(a);
    }
     
    // function to print result
    static void printNumberOfNodes()
    {
        for (int i = 1; i < N; i++)
            Console.WriteLine("Node of a subtree of "+ i +
                                        " : "+ count1[i]);
    }
     
    // Driver Code
    public static void Main(String[] args)
    {
        // Creating list for all nodes
        for(int i = 0; i < N; i++)
            adj[i] = new List<int>();
             
        // insertion of nodes in graph
        addEdge(1, 2);
        addEdge(1, 4);
        addEdge(1, 5);
        addEdge(2, 6);
        addEdge(4, 3);
        addEdge(4, 7);
         
        // call to perform dfs calculation
        // making 1 as root of tree
        numberOfNodes(1, 0);
         
        // print result
        printNumberOfNodes();
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
    // A Javascript code to find number of nodes
    // in subtree of each node
     
    // variables used to store data globally
    let N = 8;
    let count1 = new Array(N);
       
    // adjacency list representation of tree
    let adj = new Array(N);
       
    // function to calculate no. of nodes in a subtree
    function numberOfNodes(s, e)
    {
        count1[s] = 1;
        for(let u = 0; u < adj[s].length; u++)
        {
            // condition to omit reverse path
            // path from children to parent
            if(adj[s][u] == e)
                continue;
               
            // recursive call for DFS
            numberOfNodes(adj[s][u] ,s);
               
            // update count[] value of parent using
            // its children
            count1[s] += count1[adj[s][u]];
        }
    }
       
    // function to add edges in graph
    function addEdge(a, b)
    {
        adj[a].push(b);
        adj[b].push(a);
    }
       
    // function to print result
    function printNumberOfNodes()
    {
        for (let i = 1; i < N; i++)
            document.write("Node of a subtree of "+ i+
                                       " : "+ count1[i] + "</br>");
    }
     
    // Creating list for all nodes
    for(let i = 0; i < N; i++)
      adj[i] = [];
 
    // insertion of nodes in graph
    addEdge(1, 2);
    addEdge(1, 4);
    addEdge(1, 5);
    addEdge(2, 6);
    addEdge(4, 3);
    addEdge(4, 7);
 
    // call to perform dfs calculation
    // making 1  as root of tree
    numberOfNodes(1, 0);
 
    // print result
    printNumberOfNodes();
     
    // This code is contributed by suresh07.
</script>

Output: 
 

Nodes in subtree of 1: 7
Nodes in subtree of 2: 2
Nodes in subtree of 3: 1
Nodes in subtree of 4: 3
Nodes in subtree of 5: 1
Nodes in subtree of 6: 1
Nodes in subtree of 7: 1

Input and Output illustration: 
 

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