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..
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 contiguous 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 + height 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; // Adjacency 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 height under[child] += under[it]; } } } // Function to print the DFS of subtree of node 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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Improved By : ManasChhabra2