Two Sum BSTs in C++

Suppose we have two binary search trees, we have to return True iff there is a node in the first tree and a node in the second tree and sum of these nodes is a given integer target. So if the tree is like −

and target is 5, then the result is true.

To solve this, we will follow these steps −

• Define a map s
• define a method called check(), this will take node, target and nodeNumber, this will work as follows −
• if node is valid, then return false
• curr := value of node, req := target – curr
• if req is present in s and s[req] is not nodeNumber, then return true
• s[curr] := nodeNumber
• return check(left of node, target, nodeNumber) OR check(right of node, target, nodeNumber)
• In the main method, we will set −
• flag := check(root1, target, 1)
• return check(root2, target, 2)

Example(C++)

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class TreeNode{
   public:
      int val;
      TreeNode *left, *right;
      TreeNode(int data){
         val = data;
         left = NULL;
         right = NULL;
      }
};
void insert(TreeNode **root, int val){
   queue<TreeNode*> q;
   q.push(*root);
   while(q.size()){
      TreeNode *temp = q.front();
      q.pop();
      if(!temp->left){
         if(val != NULL)
            temp->left = new TreeNode(val);
         else
            temp->left = new TreeNode(0);
         return;
      }
      else{
         q.push(temp->left);
      }
      if(!temp->right){
         if(val != NULL)
            temp->right = new TreeNode(val);
         else
            temp->right = new TreeNode(0);
         return;
      }
      else{
         q.push(temp->right);
      }
   }
}
TreeNode *make_tree(vector<int> v){
   TreeNode *root = new TreeNode(v[0]);
   for(int i = 1; i<v.size(); i++){
      insert(&root, v[i]);
   }
   return root;
}
class Solution {
public:
   map <int,int> s;
   bool check(TreeNode* node, int target,int nodeNumber){
      if(!node)return false;
      int curr = node->val;
      int req = target - curr;
      if(s.find(req)!=s.end() && s[req]!=nodeNumber)return true;
      s[curr]=nodeNumber;
      return check(node->left,target,nodeNumber) || check(node->right,target,nodeNumber);
   }
   bool twoSumBSTs(TreeNode* root1, TreeNode* root2, int target) {
      bool flag = check(root1,target,1);
      return check(root2,target,2);
   }
};
main(){
   vector<int> v1 = {2,1,4};
   vector<int> v2 = {1,0,3};
   TreeNode *r1 = make_tree(v1);
   TreeNode *r2 = make_tree(v2);
   Solution ob;
   cout <<ob.twoSumBSTs(r1, r2, 5);
}

Input

[2,1,4]
[1,0,3]
5

Output

1