Queries for M-th node in the DFS of subtree
Given a tree of N nodes and N-1 edges. Also given an integer M and a node, the task is to print the M-th node in the DFS of the subtree of a given node for multiple queries.
Note: M will not be greater than the number of nodes in the subtree of the given node.
Input: M = 3, node = 1
Output: 4
In the above example if 1 is given as the node, then the DFS of subtree will be 1 2 4 6 7 5 3, hence if M is 3, then the 3rd node is 4Input: M = 4, node = 2
Output: 7
If 2 is given as the node, then the DFS of the subtree will be 2 4 6 7 5., hence if M is 4 then the 4th node is 7.
Approach:
- Add the edges between the nodes in an adjacency list.
- Call DFS function to generate the DFS of the complete tree.
- Use an 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.
- Let index of given node in the DFS of the tree be ind, then the M-th node will be at index ind + M -1 as the DFS of a subtree of a node will always be a contiguous subarray starting from the 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; // 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 height under[child] += under[it]; } } } // Function to return the DFS of subtree of node int printnodeDFSofSubtree(int node, int under[], int m) { // index of node in the original DFS int ind = mp[node]; // height of subtree of node return a[ind + m - 1]; } // 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(); int m = 3; // Query 1 cout << printnodeDFSofSubtree(1, under, m) << endl; // Query 2 m = 4; cout << printnodeDFSofSubtree(2, under, m); 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 printnodeDFSofSubtree(node, under, m): # index of node in the original DFS ind = mp[node] # height of subtree of node return a[ind + m - 1] # 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() m = 3 # Query 1 print(printnodeDFSofSubtree(1, under, m)) # Query 2 m = 4print(printnodeDFSofSubtree(2, under, m)) # This code is contributed by SHUBHAMSINGH10 |
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Output:
4 7
Time Complexity: O(1), for processing each query.
Auxiliary Space: O(N)
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