Maximum sum of nodes in Binary tree such that no two are adjacent | Dynamic Programming In C++

C++Server Side ProgrammingProgramming

In this tutorial, we will be discussing a program to find maximum sum of nodes in Binary tree such that no two are adjacent using Dynamic Programming.

For this we will be provided with a binary tree. Our task is to find the subset having maximum sum such that no two nodes in the subset are directly connected using Dynamic Programming.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
//finding diameter using dynamic programming
void dfs(int node, int parent, int dp1[], int dp2[], list<int>* adj, int tree[]){
   int sum1 = 0, sum2 = 0;
   for (auto i = adj[node].begin(); i != adj[node].end();
      ++i) {
         if (*i == parent)
            continue;
         dfs(*i, node, dp1, dp2, adj, tree);
         sum1 += dp2[*i];
         sum2 += max(dp1[*i], dp2[*i]);
      }
      dp1[node] = tree[node] + sum1;
      dp2[node] = sum2;
}
int main() {
   int n = 5;
   list<int>* adj = new list<int>[n + 1];
   adj[1].push_back(2);
   adj[2].push_back(1);
   adj[1].push_back(3);
   adj[3].push_back(1);
   adj[2].push_back(4);
   adj[4].push_back(2);
   adj[2].push_back(5);
   adj[5].push_back(2);
   int tree[n + 1];
   tree[1] = 10;
   tree[2] = 5;
   tree[3] = 11;
   tree[4] = 6;
   tree[5] = 8;
   int dp1[n + 1], dp2[n + 1];
   memset(dp1, 0, sizeof dp1);
   memset(dp2, 0, sizeof dp2);
   dfs(1, 1, dp1, dp2, adj, tree);
   cout << "Maximum sum: " << max(dp1[1], dp2[1]) << endl;
   return 0;
}

Output

Maximum sum: 25
raja
Published on 09-Sep-2020 16:38:04
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