Brothers From Different Roots

class Solution
{
public:
    int countPairs(Node* root1, Node* root2, int x)
    {
       //see as we know the condition can arise when we will be having sum i.e a+b>x || a+b<x
       //so we should be doing some stuff so that when situation comes we can increase or decrease
       
       stack<Node*>s1;
       stack<Node*>s2;
       int ans=0;
       Node*a=root1;
       Node*b=root2;
       
       while(true)
       {
           while(a)
           {
               s1.emplace(a);
               a=a->left;
           }
           
           while(b)
           {
               s2.emplace(b);
               b=b->right;
           }
           
           //if at any point the any of the stack is empty then we should break
           if(s1.empty() || s2.empty())
           {
               break;
           }
           
           auto atop=s1.top();
           auto btop=s2.top();
           
           int sum=atop->data+btop->data;
           if(sum==x)
           {
               ans++;
               s1.pop();
               s2.pop();
               a=atop->right;
               b=btop->left;
           }
           else if(sum<x)
           {
               //means we should increase the sum which implies should increase the value from that bst from which we were taking small values
               s1.pop();
               a=atop->right;
           }
           else
           {
               s2.pop();
               b=btop->left;
           }
       }
       return ans;
    }
};