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 4Output:Subtree of node 0 is 1 2 3 4 Subtree of node 1 is 2 Subtree of node 3 is 4Input:N = 7 0 1 1 2 2 3 0 4 4 5 4 6Output: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++ 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; ` ` ` `} ` |

*chevron_right*

*filter_none*

**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

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