Subtrees formed after bursting nodes

You are given an n-ary tree with a special property: If we burst a random node of the tree, this node along with its immediate parents up to the root vanishes. The tree has N nodes and nodes are numbered from 1 to N. The root is always at 1. Given a sequence of queries denoting the number of the node we start bursting, the problem is to find the number of subtrees that would be formed in the end according to the above property, for each query independently.

Examples:

Input:
Consider the following tree:
             1
           / | \ 
          2  3  4
            / \   \
           5   6   7
                  / \
                 8   9

q = 2
n = 1  
n = 7   
Output:
3
4
Explanation:
In the first query after bursting node 1, there 
will be 3 subtrees formed rooted at 2, 3 and 4.
In the second query after bursting node 7, nodes 
4 and 1 also get burst, thus there will 
be 4 subtrees formed rooted at 8, 9, 2 and 3.



Since we are dealing with n-ary tree we can use a representation similar to that of a graph, and add the bidirectional edges in an array of lists. Now if we burst a node, we can say for sure that all its children will become separate subtrees. Moreover all the children of its parents and others ancestors till the root that burst, will also become separate subtrees. So in our final answer we want to exclude the current node and all its ancestors in the path till the root. Thus we can form the equation to solve as:

answer[node] = degree[node] + allChild[parent[node]] – countPath[node]

where allChild[]: number of node’s children + number of its
parent’s children + ..+ number of root’s children
parent[]: parent of a node in the tree
degree[]: number of children for a node
countPath[]: number of nodes from root to parent of node

We can fill all the above arrays using depth first search over the adjacency list.We can start from the root 1, assuming its parent is 0 and recur depth first to propagate its values to its children. Thus we can pre-process and fill the above arrays initially and return the equation’s value for each query accordingly.

Following is the C++ implementation of the above approach:

// CPP program to find number of subtrees after bursting nodes
#include <bits/stdc++.h>
using namespace std;

// do depth first search of node nod; par is its parent
void dfs(int nod, int par, list<int> adj[], int allChild[],
         int parent[], int degree[], int countPath[])
{
    // go through the adjacent nodes
    for (auto it = adj[nod].begin(); it != adj[nod].end(); it++) {
        int curr = *it;

        // avoid cycling
        if (curr == par)
            continue;

        degree[nod]++;
        countPath[curr] = countPath[nod] + 1;
        parent[curr] = nod;
    }

    // propagated from parent
    allChild[nod] = allChild[parent[nod]] + degree[nod];

    // go through the adjacent nodes
    for (auto it = adj[nod].begin(); it != adj[nod].end(); it++) {
        int curr = *it;

        // avoid cycling
        if (curr == par)
            continue;

        // recur and go depth first
        dfs(curr, nod, adj, allChild, parent, degree, countPath);
    }
}

// Driver code
int main()
{
    int n = 9;

    // adjacency list for each node
    list<int> adj[n + 1];

    // allChild[]: number of node's children + number of its
    // parent's children + ..+ number of root's children
    // parent[]: parent of a node in the tree
    // degree[]: number of children for a node
    // countPath[]: number of nodes from root to parent of node
    int allChild[n + 1] = { 0 }, parent[n + 1] = { 0 }, 
       degree[n + 1] = { 0 }, countPath[n + 1] = { 0 };

    // construct tree
    adj[1].push_back(2);
    adj[2].push_back(1);
    adj[1].push_back(3);
    adj[3].push_back(1);
    adj[1].push_back(4);
    adj[4].push_back(1);
    adj[3].push_back(5);
    adj[5].push_back(3);
    adj[3].push_back(6);
    adj[6].push_back(3);
    adj[4].push_back(7);
    adj[7].push_back(4);
    adj[7].push_back(8);
    adj[8].push_back(7);
    adj[7].push_back(9);
    adj[9].push_back(7);

    // assume 1 is root and 0 is its parent
    dfs(1, 0, adj, allChild, parent, degree, countPath);

    // 2 queries
    int curr = 1;
    cout << degree[curr] + allChild[parent[curr]] - countPath[curr] << endl;

    curr = 7;
    cout << degree[curr] + allChild[parent[curr]] - countPath[curr] << endl;

    return 0;
}

Output:
3
4

The time complexity of the above algorithm is O(E * lg(V)) where E id the number of edges and
V is the number of vertices.



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