Frog Position After T Seconds.java

class Solution {
    public double frogPosition(int n, int[][] edges, int t, int target) {
       
        List<List<Integer>> graph=new ArrayList<>();
        for(int i=0;i<=n;i++) graph.add(new ArrayList<>());
        
          for(int i=0;i<edges.length;i++) {
         
           graph.get(edges[i][0]).add(edges[i][1]);
           graph.get(edges[i][1]).add(edges[i][0]);
          
      }
        boolean[] vis=new boolean[n+1];
        return Sol(graph,1, t, target,vis);
        
    }
    public double Sol(List<List<Integer>> graph,int ver,int t,int tar,boolean[] vis){
        
         int count=0;
        for(Integer child:graph.get(ver)){
            if(!vis[child]) count++;
        }
        
        
        vis[ver]=true;
        if(t<0) return 0;
        
        if(ver==tar){
            if(count==0 || t==0) return 1.0;
         }
        
        if(graph.get(ver).size()==0) return 0;
        
        double ans=0.0;
        
       
        for(Integer child:graph.get(ver)){

           // System.out.println(child);
            if(!vis[child])
            ans+=(double)(1.0/count)*Sol(graph,child,t-1,tar,vis);
        }
       // System.out.println(ans);
        vis[ver]=false;
        return ans;
    }
}