Subtree of all nodes in a tree using DFS
Last Updated :
03 Nov, 2022
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:
- When function dfs(0, 0) is called, start[0] = 0, dfs_order.push_back(0), visited[0] = 1 to keep track of dfs order.
- Now, consider adjacency list (adj[100001]) as considering directional path elements connected to node 0 will be in adjacency list corresponding to node 0.
- Now, recursively call dfs function till all elements traversed of adj[0].
- Now, dfs(1, 2) is called, Now start[1] = 1, dfs_order.push_back(1), visited[1] = 1 after adj[1] elements is traversed.
- 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.
- 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 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 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# 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.
<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)