# Subtree of all nodes in a tree using DFS

• Difficulty Level : Easy
• Last Updated : 29 Oct, 2021

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.

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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``using` `namespace` `std;` `// arrays for keeping position``// at each dfs traversal for each node``int` `start[100001];``int` `end[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 "``<

## 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 dfs_order = ``new` `Vector();``    ``static` `Vector> adj = ``new` `Vector>();``    ``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());``        ``}``          ` `        ``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> adj = ``new` `List>();``    ``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

 ``
Output:
```subtree of node 0 is 1 4 7 8 5 2 6 9 3
subtree of node 1 is 4 7 8 5
subtree of node 2 is 6 9
subtree of node 4 is 7 8
subtree of node 6 is 9```

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

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