Count Nodes Whose Sum with X is a Fibonacci Number in C++

Given a binary tree with weights of its nodes as numbers. The goal is to find the number of nodes that have weights such that the number is a Fibonacci number. Numbers in Fibonacci series are: 0, 1, 1, 2, 3, 5, 8, 13….nth number is the sum of (n−1)th and (n−2)th. If weight is 13 then it is a Fibonacci number so the node will be counted.

For Example

Input

temp =1. The tree which will be created after inputting the values is given below −

Output

Count the nodes whose sum with X is a Fibonacci number are: 3

Explanation

we are given with the tree nodes and the weights associated with each
node. Now we check whether the temp+weight is a Fibonacci number or not.

Input

temp = 3. The tree which will be created after inputting the values is given below −

Output

Count the nodes whose sum with X is a Fibonacci number are: 3

Explanation

we are given with the tree nodes and the weights associated with each
node. Now we check whether the temp+weight is a Fibonacci number or not.

Approach used in the below program is as follows −

In this approach we will apply DFS on the graph of the tree to traverse it and check if the weights of the node and temp add upto a fibonacci number. Take two vectors Node_Weight(100) and edge_graph[100] for this purpose.

• Initialize Node_Weight[] with the weights of nodes.

• Create a tree using vector edge_graph.

• Take a global variable Fibonacci and initialize it with 0. Take other global variable temp.

• Function check_square(long double val) takes an integer and returns true if val is a perfect square.

• Take val_1 = sqrt(val)

• Now if(val_1 − floor(val_1) == 0) returns true then total is a perfect square, return true.

• Return false otherwise.

• Function check_Fibonacci(int num) takes a number and returns true if it is a fibonacci number.

• Initialize fib with 5*num*num.

• If check_square((fib + 4)) || check_square((fib − 4)) results true then return true.

• Return false otherwise.

• Function Fibonacci_number(int node, int root) returns count of nodes whose sum with X is a Fibonacci number.

• If if(check_Fibonacci(Node_Weight[node] + temp)) returns true then increment Fibonacci.

• Traverse tree in vector edge_graph[node] using for loop.

• Call Fibonacci_number(it, node) for the next node in the vector.

• At the end of all functions we will have a Fibonacci as the number of nodes with weights having sum with temp as a fibonacci number.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
vector<int> Node_Weight(100);
vector<int> edge_graph[100];
int Fibonacci = 0, temp;
bool check_square(long double val){
   long double val_1 = sqrt(val);
   if(val_1 − floor(val_1) == 0){
      return true;
   }
   return false;
}
bool check_Fibonacci(int num){
   int fib = 5 * num * num;
   if(check_square((fib + 4)) || check_square((fib − 4))){
      return true;
   }
   return false;
}
void Fibonacci_number(int node, int root){
   if(check_Fibonacci(Node_Weight[node] + temp)){
      Fibonacci++;
   }
   for (int it : edge_graph[node]){
      if(it == root){
         continue;
      }
      Fibonacci_number(it, node);
   }
}
int main(){
   //weight of the nodes
   Node_Weight[2] = 6;
   Node_Weight[1] = 4;
   Node_Weight[4] = 23;
   Node_Weight[3] = 5;
   Node_Weight[8] = 161;
   Node_Weight[9] = 434;
   //create graph edge
   edge_graph[2].push_back(1);
   edge_graph[2].push_back(4);
   edge_graph[4].push_back(3);
   edge_graph[4].push_back(8);
   edge_graph[8].push_back(9);
   temp = 3;
   Fibonacci_number(2, 2);
   cout<<"Count the nodes whose sum with X is a Fibonacci number are: "<<Fibonacci;
   return 0;
}

Output

If we run the above code it will generate the following output −

Count the nodes whose sum with X is a Fibonacci number are: 1