Count the nodes in the given Tree whose weight is a Perfect Number | Java implementation to Count the nodes in the given tree whose weight is a Perfect Number ; Function that returns true if n is perfect ; Variable to store sum of divisors ; Find all divisors and add them ; Check if sum of divisors is equal to n , then n is a perfect number ; Function to perform dfs ; If weight of the current node is a perfect number ; Driver code ; Weights of the node ; Edges of the tree
import java . util . * ; class GFG { static int ans = 0 ; static Vector < Integer > [ ] graph = new Vector [ 100 ] ; static int [ ] weight = new int [ 100 ] ; static boolean isPerfect ( int n ) { int sum = 1 ; for ( int i = 2 ; i * i <= n ; i ++ ) { if ( n % i == 0 ) { if ( i * i != n ) sum = sum + i + n / i ; else sum = sum + i ; } } if ( sum == n && n != 1 ) return true ; return false ; } static void dfs ( int node , int parent ) { if ( isPerfect ( weight [ node ] ) ) ans += 1 ; for ( int to : graph [ node ] ) { if ( to == parent ) continue ; dfs ( to , node ) ; } } public static void main ( String [ ] args ) { for ( int i = 0 ; i < graph . length ; i ++ ) graph [ i ] = new Vector < Integer > ( ) ; weight [ 1 ] = 5 ; weight [ 2 ] = 10 ; weight [ 3 ] = 11 ; weight [ 4 ] = 8 ; weight [ 5 ] = 6 ; graph [ 1 ] . add ( 2 ) ; graph [ 2 ] . add ( 3 ) ; graph [ 2 ] . add ( 4 ) ; graph [ 1 ] . add ( 5 ) ; dfs ( 1 , 1 ) ; System . out . print ( ans ) ; } }