# Print all pairs from two BSTs whose sum is greater than the given value

Given two Binary Search Tree (BSTs) and a value X, the problem is to print all pairs from both the BSTs whose sum is greater than the given value X.

Examples:

```Input:
BST 1:
5
/   \
3     7
/ \   / \
2  4  6   8
BST 2:
10
/   \
6     15
/ \   /  \
3  8  11  18
X = 20
Output: The pairs are:
(3, 18)
(4, 18)
(5, 18)
(6, 18)
(7, 18)
(8, 18)
(6, 15)
(7, 15)
(8, 15)
```

Naive Approach: For each node value A in BST 1, search the value in BST 2 which is greater than the (X – A). If the value is found then print the pair.

Time complexity: O(n1 * h2), where n1 is the number of nodes in first BST and h2 is the height of the second BST.

Efficient Approach:

1. Traverse BST 1 from smallest value to node to largest by taking index i. This can be achieved with the help of inorder traversal.
2. Traverse BST 2 from largest value node to smallest by taking index j. This can be achieved with the help of inorder traversal.
3. Perform these two traversals one by one and store into two array.
4. Sum up the corresponding node’s value from both the BSTs at a particular instance of traversals.

• If sum > x, then print pair and decrement j by 1.
• If x > sum, then increment i by 1.

Perform these operations until either of the two traversals gets completed.

Below is the implementation of the above approach:

 `// C++ implementation to print pairs ` `// from two BSTs whose sum is greater ` `// the given value x ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Structure of each node of BST ` `struct` `node { ` `    ``int` `key; ` `    ``struct` `node *left, *right; ` `}; ` ` `  `// Function to create a new BST node ` `node* newNode(``int` `item) ` `{ ` `    ``node* temp = ``new` `node(); ` `    ``temp->key = item; ` `    ``temp->left = temp->right = NULL; ` `    ``return` `temp; ` `} ` ` `  `// A utility function to insert a ` `// new node with given key in BST ` `struct` `node* insert(``struct` `node* node, ` `                    ``int` `key) ` `{ ` `    ``// If the tree is empty, return a new node ` `    ``if` `(node == NULL) ` `        ``return` `newNode(key); ` ` `  `    ``// Otherwise, recur down the tree ` `    ``if` `(key < node->key) ` `        ``node->left = insert(node->left, ` `                            ``key); ` `    ``else` `if` `(key > node->key) ` `        ``node->right = insert(node->right, ` `                             ``key); ` ` `  `    ``// Return the (unchanged) node pointer ` `    ``return` `node; ` `} ` ` `  `// Function to return the size of ` `// the tree ` `int` `sizeOfTree(node* root) ` `{ ` `    ``if` `(root == NULL) { ` `        ``return` `0; ` `    ``} ` ` `  `    ``// Calculate left size recursively ` `    ``int` `left = sizeOfTree(root->left); ` ` `  `    ``// Calculate right size recursively ` `    ``int` `right = sizeOfTree(root->right); ` ` `  `    ``// Return total size recursively ` `    ``return` `(left + right + 1); ` `} ` ` `  `// Function to store inorder ` `// traversal of BST ` `void` `storeInorder(node* root, ` `                  ``int` `inOrder[], ` `                  ``int``& index) ` `{ ` `    ``// Base condition ` `    ``if` `(root == NULL) { ` `        ``return``; ` `    ``} ` ` `  `    ``// Left recursive call ` `    ``storeInorder(root->left, ` `                 ``inOrder, ` `                 ``index); ` ` `  `    ``// Store elements in inorder array ` `    ``inOrder[index++] = root->key; ` ` `  `    ``// Right recursive call ` `    ``storeInorder(root->right, ` `                 ``inOrder, ` `                 ``index); ` `} ` ` `  `// Function to print the pairs ` `void` `print(``int` `inOrder1[], ``int` `i, ` `           ``int` `index1, ``int` `value) ` `{ ` `    ``while` `(i < index1) { ` `        ``cout << ``"("` `<< inOrder1[i] ` `             ``<< ``", "` `<< value ` `             ``<< ``")"` `<< endl; ` `        ``i++; ` `    ``} ` `} ` ` `  `// Utility function to check the ` `// pair of BSTs whose sum is ` `// greater than given value x ` `void` `printPairUtil(``int` `inOrder1[], ` `                   ``int` `inOrder2[], ` `                   ``int` `index1, ` `                   ``int` `j, ``int` `k) ` `{ ` `    ``int` `i = 0; ` ` `  `    ``while` `(i < index1 && j >= 0) { ` ` `  `        ``if` `(inOrder1[i] + inOrder2[j] > k) { ` `            ``print(inOrder1, i, ` `                  ``index1, inOrder2[j]); ` `            ``j--; ` `        ``} ` `        ``else` `{ ` `            ``i++; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to check the ` `// pair of BSTs whose sum is ` `// greater than given value x ` `void` `printPairs(node* root1, ` `                ``node* root2, ``int` `k) ` `{ ` `    ``// Store the size of BST1 ` `    ``int` `numNode = sizeOfTree(root1); ` ` `  `    ``// Take auxiliary array for storing ` `    ``// The inorder traversal of BST1 ` `    ``int` `inOrder1[numNode + 1]; ` `    ``int` `index1 = 0; ` ` `  `    ``// Store the size of BST2 ` `    ``numNode = sizeOfTree(root2); ` ` `  `    ``// Take auxiliary array for storing ` `    ``// The inorder traversal of BST2 ` `    ``int` `inOrder2[numNode + 1]; ` `    ``int` `index2 = 0; ` ` `  `    ``// Function call for storing ` `    ``// inorder traversal of BST1 ` `    ``storeInorder(root1, inOrder1, ` `                 ``index1); ` ` `  `    ``// Function call for storing ` `    ``// inorder traversal of BST1 ` `    ``storeInorder(root2, inOrder2, ` `                 ``index2); ` ` `  `    ``// Utility function call to count ` `    ``// the pair ` `    ``printPairUtil(inOrder1, inOrder2, ` `                  ``index1, index2 - 1, k); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``/* Formation of BST 1  ` `             ``5 ` `           ``/   \       ` `          ``3     7      ` `         ``/ \   / \     ` `         ``2  4  6  8   ` `    ``*/` ` `  `    ``struct` `node* root1 = NULL; ` `    ``root1 = insert(root1, 5); ` `    ``insert(root1, 3); ` `    ``insert(root1, 2); ` `    ``insert(root1, 4); ` `    ``insert(root1, 7); ` `    ``insert(root1, 6); ` `    ``insert(root1, 8); ` ` `  `    ``/* Formation of BST 2  ` `            ``10 ` `           ``/   \       ` `          ``6     15      ` `         ``/ \   / \     ` `        ``3   8 11  18   ` `    ``*/` ` `  `    ``struct` `node* root2 = NULL; ` `    ``root2 = insert(root2, 10); ` `    ``insert(root2, 6); ` `    ``insert(root2, 15); ` `    ``insert(root2, 3); ` `    ``insert(root2, 8); ` `    ``insert(root2, 11); ` `    ``insert(root2, 18); ` ` `  `    ``int` `x = 20; ` ` `  `    ``// Print pairs ` `    ``printPairs(root1, root2, x); ` ` `  `    ``return` `0; ` `} `

Output:

```(3, 18)
(4, 18)
(5, 18)
(6, 18)
(7, 18)
(8, 18)
(6, 15)
(7, 15)
(8, 15)
```

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