Count of subtrees from an N-ary tree consisting of single colored nodes

Given a N-ary tree consisting of N nodes and a matrix edges[][] consisting of N – 1 edges of the form (X, Y) denoting the edge between node X and node Y and an array col[] consisting of values:

  • 0: Uncolored node.
  • 1: Node colored red.
  • 2: Node colored blue.

The task is to find the number of subtrees of the given tree which consists of only single-colored nodes.

Examples:

Input:
N = 5, col[] = {2, 0, 0, 1, 2},
edges[][] = {{1, 2}, {2, 3}, {2, 4}, {2, 5}}
Output: 1
Explanation:
A subtree of node 4 which is {4} has no blue vertex and contains only one red vertex.

Input:
N = 5, col[] = {1, 0, 0, 0, 2},
edges[][] = {{1, 2}, {2, 3}, {3, 4}, {4, 5}}
Output: 4
Explanation:
Below are the subtrees with the given property:



  1. Subtree with root node value 2 {2, 3, 4, 5}
  2. Subtree with root node value 3 {3, 4, 5}
  3. Subtree with root node value 4 {4, 5}
  4. Subtree with root node value 5 {5}

Approach: The given problem can be solved using Depth First Search Traversal. The idea is to calculate the number of red and blue nodes in each subtree using DFS for the given tree. Once calculated, count the number of subtrees containing only blue colored nodes and only red colored nodes.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to implement DFS traversal
void Solution_dfs(int v, int color[], int red,
                  int blue, int* sub_red,
                  int* sub_blue, int* vis,
                  map<int, vector<int> >& adj,
                  int* ans)
{
  
    // Mark node v as visited
    vis[v] = 1;
  
    // Traverse Adj_List of node v
    for (int i = 0; i < adj[v].size();
         i++) {
  
        // If current node is not visited
        if (vis[adj[v][i]] == 0) {
  
            // DFS call for current node
            Solution_dfs(adj[v][i], color,
                         red, blue,
                         sub_red, sub_blue,
                         vis, adj, ans);
  
            // Count the total red and blue
            // nodes of children of its subtree
            sub_red[v] += sub_red[adj[v][i]];
            sub_blue[v] += sub_blue[adj[v][i]];
        }
    }
  
    if (color[v] == 1) {
        sub_red[v]++;
    }
  
    // Count the no. of red and blue
    // nodes in the subtree
    if (color[v] == 2) {
        sub_blue[v]++;
    }
  
    // If subtree contains all
    // red node & no blue node
    if (sub_red[v] == red
        && sub_blue[v] == 0) {
        (*ans)++;
    }
  
    // If subtree contains all
    // blue node & no red node
    if (sub_red[v] == 0
        && sub_blue[v] == blue) {
        (*ans)++;
    }
}
  
// Function to count the number of
// nodes with red color
int countRed(int color[], int n)
{
    int red = 0;
    for (int i = 0; i < n; i++) {
        if (color[i] == 1)
            red++;
    }
    return red;
}
  
// Function to count the number of
// nodes with blue color
int countBlue(int color[], int n)
{
    int blue = 0;
    for (int i = 0; i < n; i++) {
        if (color[i] == 2)
            blue++;
    }
    return blue;
}
  
// Function to create a Tree with
// given vertices
void buildTree(int edge[][2],
               map<int, vector<int> >& m,
               int n)
{
    int u, v, i;
  
    // Traverse the edge[] array
    for (i = 0; i < n - 1; i++) {
  
        u = edge[i][0] - 1;
        v = edge[i][1] - 1;
  
        // Create adjacency list
        m[u].push_back(v);
        m[v].push_back(u);
    }
}
  
// Function to count the number of
// subtree with the given condition
void countSubtree(int color[], int n,
                  int edge[][2])
{
  
    // For creating adjacency list
    map<int, vector<int> > adj;
    int ans = 0;
  
    // To store the count of subtree
    // with only blue and red color
    int sub_red[n + 3] = { 0 };
    int sub_blue[n + 3] = { 0 };
  
    // visted array for DFS Traversal
    int vis[n + 3] = { 0 };
  
    // Count the number of red
    // node in the tree
    int red = countRed(color, n);
  
    // Count the number of blue
    // node in the tree
    int blue = countBlue(color, n);
  
    // Function Call to build tree
    buildTree(edge, adj, n);
  
    // DFS Traversal
    Solution_dfs(0, color, red, blue,
                 sub_red, sub_blue,
                 vis, adj, &ans);
  
    // Print the final count
    cout << ans;
}
// Driver Code
int main()
{
    int N = 5;
    int color[] = { 1, 0, 0, 0, 2 };
    int edge[][2] = { { 1, 2 },
                      { 2, 3 },
                      { 3, 4 },
                      { 4, 5 } };
  
    countSubtree(color, N, edge);
  
    return 0;
}

chevron_right


Output:

4

Time Complexity: O(N)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.