Queries for DFS of a subtree in a tree

Given a tree of N nodes and N-1 edges. The task is to print the DFS of the subtree of a given node for multiple queries. The DFS must include the given node as the root of the subtree.

In the above tree, if 1 is given as the node, then the DFS of subtree will be 1 2 4 6 7 5 3.

If 2 is given as the node, then the DFS of the subtree will be 2 4 6 7 5..

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Add the edges between the nodes in an adjacency list.
Call DFS function to generate the DFS of the complete tree.
Use a under[] array to store the height of the subtree under the given node including the node.
In the DFS function, keep incrementing the size of subtree on every recursive call.
Mark the node index in the DFS of complete using hashing.
The DFS of a subtree of a node will always be a contigous subarray starting from the node(say index ind) to (ind+height of subtree).
Get the index of node which has been stored using hashing and print the nodes from original DFS till index = ind + heigth of subtree which has been stored in under[node].

Below is the implementation of the above approach.

// C++ program for Queries 
// for DFS of subtree of a node in a tree 
#include <bits/stdc++.h> 
using namespace std; 
const int N = 100000; 
  
// Adjaceny list to store the 
// tree nodes connection 
vector<int> v[N]; 
  
// stores the index of node in DFS 
unordered_map<int, int> mp; 
  
// stores the index of node in 
// original node 
vector<int> a; 
  
// Function to call DFS and count nodes 
// under that subtree 
void dfs(int under[], int child, int parent) 
{ 
  
    // stores the DFS of tree 
    a.push_back(child); 
  
    // hieght of subtree 
    under[child] = 1; 
  
    // iterate for children 
    for (auto it : v[child]) { 
  
        // if not equal to parent 
        // so that it does not traverse back 
        if (it != parent) { 
  
            // call DFS for subtree 
            dfs(under, it, child); 
  
            // add the heigth 
            under[child] += under[it]; 
        } 
    } 
} 
  
// Function to print the DFS of subtree of nodec 
void printDFSofSubtree(int node, int under[]) 
{ 
    // index of node in the original DFS 
    int ind = mp[node]; 
  
    // height of subtree of node 
    int height = under[node]; 
  
    cout << "The DFS of subtree " << node << ": "; 
  
    // print the DFS of subtree 
    for (int i = ind; i < ind + under[node]; i++) { 
        cout << a[i] << " "; 
    } 
    cout << endl; 
} 
  
// Function to add edges to a tree 
void addEdge(int x, int y) 
{ 
    v[x].push_back(y); 
    v[y].push_back(x); 
} 
  
// Marks the index of node in original DFS 
void markIndexDfs() 
{ 
    int size = a.size(); 
  
    // marks the index 
    for (int i = 0; i < size; i++) { 
        mp[a[i]] = i; 
    } 
} 
  
// Driver Code 
int main() 
{ 
    int n = 7; 
  
    // add edges of a tree 
    addEdge(1, 2); 
    addEdge(1, 3); 
    addEdge(2, 4); 
    addEdge(2, 5); 
    addEdge(4, 6); 
    addEdge(4, 7); 
  
    // array to store the height of subtree 
    // of every node in a tree 
    int under[n + 1]; 
  
    // Call the function DFS to generate the DFS 
    dfs(under, 1, 0); 
  
    // Function call to mark the index of node 
    markIndexDfs(); 
  
    // Query 1 
    printDFSofSubtree(2, under); 
  
    // Query 1 
    printDFSofSubtree(4, under); 
  
    return 0; 
} 

Time Complexity: O( N + M ), where N is the number of nodes and M is the number of edges for pre-calculation and O(N) for queries in worst case.
Auxiliary Space: O(N)

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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