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++
// 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; } |
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Python3
# Python3 program for Queries # for DFS of subtree of a node in a tree N = 100000 # Adjaceny list to store the # tree nodes connection v = [[]for i in range(N)] # stores the index of node in DFS mp = {} # stores the index of node in # original node a = [] # Function to call DFS and count nodes # under that subtree def dfs(under, child, parent): # stores the DFS of tree a.append(child) # hieght of subtree under[child] = 1 # iterate for children for it in 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 return the DFS of subtree of node def printDFSofSubtree(node, under): # index of node in the original DFS ind = mp[node] # height of subtree of node height = under[node] print("The DFS of subtree", node, ":", end=" ") # print the DFS of subtree for i in range(ind,ind + under[node]): print(a[i], end=" ") print() # Function to add edges to a tree def addEdge(x, y): v[x].append(y) v[y].append(x) # Marks the index of node in original DFS def markIndexDfs(): size = len(a) # marks the index for i in range(size): mp[a[i]] = i # Driver Code 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 under = [0]*(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 2 printDFSofSubtree(4, under) # This code is contributed by SHUBHAMSINGH10 |
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Output:
The DFS of subtree 2: 2 4 6 7 5 The DFS of subtree 4: 4 6 7
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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