Subtree of all nodes in a tree using DFS

Given n nodes of a tree and their connections, print Subtree nodes of every node.
Subtree of a node is defined as a tree which is a child of a node. The name emphasizes that everything which is a descendant of a tree node is a tree too, and is a subset of the larger tree.

Examples :

Input: N = 5
  0 1
  1 2
  0 3
  3 4
Output: 
Subtree of node 0 is 1 2 3 4 
Subtree of node 1 is 2 
Subtree of node 3 is 4

Input: N = 7
  0 1
  1 2
  2 3
  0 4
  4 5
  4 6
Output:
Subtree of node 0 is 1 2 3 4 5 6 
Subtree of node 1 is 2 3 
Subtree of node 4 is 5 6 

Approach: Do DFS traversal for every node and print all the nodes which are reachable from a particular node. 
Explanation of below code: 

1. When function dfs(0, 0) is called, start[0] = 0, dfs_order.push_back(0), visited[0] = 1 to keep track of dfs order.
2. Now, consider adjacency list (adj[100001]) as considering directional path elements connected to node 0 will be in adjacency list corresponding to node 0.
3. Now, recursively call dfs function till all elements traversed of adj[0].
4. Now, dfs(1, 2) is called, Now start[1] = 1, dfs_order.push_back(1), visited[1] = 1 after adj[1] elements is traversed.
5. Now adj [1] is traversed which contain only node 2 when adj[2] is traversed it contains no element, it will break and end[1]=2.
6. Similarly, all nodes traversed and store dfs_order in array to find subtree of nodes.

C++

// C++ code to print subtree of all nodes
#include <bits/stdc++.h>
using namespace std;

// arrays for keeping position
// at each dfs traversal for each node
int start[100001];
int endd[100001];

// Storing dfs order
vector<int> dfs_order;
vector<int> adj[100001];
int visited[100001];

// Recursive function for dfs
// traversal dfsUtil()
void dfs(int a, int& b)
{

    // keep track of node visited
    visited[a] = 1;
    b++;
    start[a] = b;
    dfs_order.push_back(a);

    for (vector<int>::iterator it = adj[a].begin();
         it != adj[a].end(); it++) {
        if (!visited[*it]) {
            dfs(*it, b);
        }
    }
    endd[a] = b;
}

// Function to print the subtree nodes
void Print(int n)
{
    for (int i = 0; i < n; i++) {
        // if node is leaf node
        // start[i] is equals to endd[i]
        if (start[i] != endd[i]) {
            cout << "subtree of node " << i << " is ";
            for (int j = start[i] + 1; j <= endd[i]; j++) {
                cout << dfs_order[j - 1] << " ";
            }
            cout << endl;
        }
    }
}

// Driver code
int main()
{
    // No of nodes n = 10
    int n = 10, c = 0;

    adj[0].push_back(1);
    adj[0].push_back(2);
    adj[0].push_back(3);
    adj[1].push_back(4);
    adj[1].push_back(5);
    adj[4].push_back(7);
    adj[4].push_back(8);
    adj[2].push_back(6);
    adj[6].push_back(9);

    // Calling dfs for node 0
    // Considering root node at 0
    dfs(0, c);

    // Print child nodes
    Print(n);

    return 0;
}

Java

// Java code to print subtree of all nodes
import java.util.*;
public class Main {
    // arrays for keeping position
    // at each dfs traversal for each node
    static int[] start = new int[100001];
    static int[] end = new int[100001];

    // Storing dfs order
    static Vector<Integer> dfs_order
        = new Vector<Integer>();
    static Vector<Vector<Integer> > adj
        = new Vector<Vector<Integer> >();
    static boolean[] visited = new boolean[100001];

    // Recursive function for dfs traversal dfsUtil()
    static int dfs(int a, int b)
    {
        // keep track of node visited
        visited[a] = true;
        b += 1;
        start[a] = b;
        dfs_order.add(a);

        for (int it = 0; it < adj.get(a).size(); it++) {
            if (!visited[adj.get(a).get(it)])
                b = dfs(adj.get(a).get(it), b);
        }

        endd[a] = b;
        return b;
    }

    // Function to print the subtree nodes
    static void Print(int n)
    {
        for (int i = 0; i < n; i++) {
            // If node is leaf node
            // start[i] is equals to endd[i]
            if (start[i] != endd[i]) {
                System.out.print("subtree of node " + i
                                 + " is ");
                for (int j = start[i] + 1; j < endd[i] + 1;
                     j++) {
                    System.out.print(dfs_order.get(j - 1)
                                     + " ");
                }
                System.out.println();
            }
        }
    }

    public static void main(String[] args)
    {
        // No of nodes n = 10
        int n = 10, c = 0;

        for (int i = 0; i < 100001; i++) {
            adj.add(new Vector<Integer>());
        }

        adj.get(0).add(1);
        adj.get(0).add(2);
        adj.get(0).add(3);
        adj.get(1).add(4);
        adj.get(1).add(5);
        adj.get(4).add(7);
        adj.get(4).add(8);
        adj.get(2).add(6);
        adj.get(6).add(9);

        // Calling dfs for node 0
        // Considering root node at 0
        dfs(0, c);

        // Print child nodes
        Print(n);
    }
}

// This code is contributed by divyeshrabadiya07.

Python3

# Python3 code to print subtree of all nodes

# arrays for keeping position at
# each dfs traversal for each node
start = [None] * 100001
end = [None] * 100001

# Storing dfs order
dfs_order = []
adj = [[] for i in range(100001)]
visited = [False] * 100001

# Recursive function for dfs traversal dfsUtil()


def dfs(a, b):

    # keep track of node visited
    visited[a] = 1
    b += 1
    start[a] = b
    dfs_order.append(a)

    for it in adj[a]:
        if not visited[it]:
            b = dfs(it, b)

    endd[a] = b
    return b

# Function to print the subtree nodes


def Print(n):

    for i in range(0, n):

        # If node is leaf node
        # start[i] is equals to endd[i]
        if start[i] != endd[i]:

            print("subtree of node", i, "is", end=" ")
            for j in range(start[i]+1, endd[i]+1):

                print(dfs_order[j-1], end=" ")

            print()


# Driver code
if __name__ == "__main__":

    # No of nodes n = 10
    n, c = 10, 0

    adj[0].append(1)
    adj[0].append(2)
    adj[0].append(3)
    adj[1].append(4)
    adj[1].append(5)
    adj[4].append(7)
    adj[4].append(8)
    adj[2].append(6)
    adj[6].append(9)

    # Calling dfs for node 0
    # Considering root node at 0
    dfs(0, c)

    # Print child nodes
    Print(n)

# This code is contributed by Rituraj Jain

C#

// C# code to print subtree of all nodes
using System;
using System.Collections.Generic;
class GFG {

    // arrays for keeping position
    // at each dfs traversal for each node
    static int[] start = new int[100001];
    static int[] end = new int[100001];

    // Storing dfs order
    static List<int> dfs_order = new List<int>();
    static List<List<int> > adj = new List<List<int> >();
    static bool[] visited = new bool[100001];

    // Recursive function for dfs traversal dfsUtil()
    static int dfs(int a, int b)
    {
        // keep track of node visited
        visited[a] = true;
        b += 1;
        start[a] = b;
        dfs_order.Add(a);

        for (int it = 0; it < adj[a].Count; it++) {
            if (!visited[adj[a][it]])
                b = dfs(adj[a][it], b);
        }

        endd[a] = b;
        return b;
    }

    // Function to print the subtree nodes
    static void Print(int n)
    {
        for (int i = 0; i < n; i++) {
            // If node is leaf node
            // start[i] is equals to endd[i]
            if (start[i] != endd[i]) {
                Console.Write("subtree of node " + i
                              + " is ");
                for (int j = start[i] + 1; j < endd[i] + 1;
                     j++) {
                    Console.Write(dfs_order[j - 1] + " ");
                }
                Console.WriteLine();
            }
        }
    }

    static void Main()
    {
        // No of nodes n = 10
        int n = 10, c = 0;

        for (int i = 0; i < 100001; i++) {
            adj.Add(new List<int>());
        }

        adj[0].Add(1);
        adj[0].Add(2);
        adj[0].Add(3);
        adj[1].Add(4);
        adj[1].Add(5);
        adj[4].Add(7);
        adj[4].Add(8);
        adj[2].Add(6);
        adj[6].Add(9);

        // Calling dfs for node 0
        // Considering root node at 0
        dfs(0, c);

        // Print child nodes
        Print(n);
    }
}

// This code is contributed by divyesh072019.

Javascript

<script>
    // Javascript code to print subtree of all nodes

    // arrays for keeping position
    // at each dfs traversal for each node
    let start = new Array(100001);
    let end = new Array(100001);

    // Storing dfs order
    let dfs_order = [];
    let adj = [];
    for(let i = 0; i < 100001; i++)
    {
        adj.push([]);
    }
    let visited = new Array(100001);
    visited.fill(false);

    // Recursive function for dfs traversal dfsUtil()
    function dfs(a, b)
    {
        // keep track of node visited
        visited[a] = true;
        b += 1;
        start[a] = b;
        dfs_order.push(a);

        for(let it = 0; it < adj[a].length; it++)
        {
            if(!visited[adj[a][it]])
                b = dfs(adj[a][it], b);
        }

        endd[a] = b;
        return b;
    }

    // Function to print the subtree nodes
    function Print(n)
    {
        for(let i = 0; i < n; i++)
        {
            // If node is leaf node
            // start[i] is equals to endd[i]
            if(start[i] != endd[i])
            {
                document.write("subtree of node "+ i + " is ");
                for(let j = start[i]+1; j < endd[i]+1; j++)
                {
                    document.write(dfs_order[j-1] +  " ");
                }
                document.write("</br>");
            }
        }
    }

    // No of nodes n = 10
    let n = 10, c = 0;

    adj[0].push(1);
    adj[0].push(2);
    adj[0].push(3);
    adj[1].push(4);
    adj[1].push(5);
    adj[4].push(7);
    adj[4].push(8);
    adj[2].push(6);
    adj[6].push(9);

    // Calling dfs for node 0
    // Considering root node at 0
    dfs(0, c);

    // Print child nodes
    Print(n);

// This code is contributed by suresh07.
</script>

Time Complexity: O(N ^ 2)
Auxiliary Space: O(N)