Count pairs from two BSTs whose sum is equal to a given value x

Given two BSTs containing n1 and n2 distinct nodes respectively. Given a value x. The problem is to count all pairs from both the BSTs whose sum is equal to x.

Examples:

Input : BST 1:    5
/   \
3     7
/ \   / \
2  4  6   8

BST 2:    10
/   \
6     15
/ \   /  \
3  8  11  18
x = 16

Output : 3
The pairs are:
(5, 11), (6, 10) and (8, 8)

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1: For each node value a in BST 1, search the value (x – a) in BST 2. If value found then increment the count. For searching a value in BST, refer this post.
Time complexity: O(n1 * h2), here n1 is number of nodes in first BST and h2 is height of second BST.

Method 2: Traverse BST 1 from smallest value to node to largest. This can be achieved with the help of iterative inorder traversal. Traverse BST 2 from largest value node to smallest. This can be achieved with the help of reverse inorder traversal. Perform these two traversals simultaneously. Sum up the corresponding node’s value from both the BSTs at a particular instance of traversals. If sum == x, then increment count. If x > sum, then move to the inorder successor of the current node of BST 1, else move to the inorder predecessor of the current node of BST 2. Perform these operations until either of the two traversals gets completed.

C++

// C++ implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
#include <bits/stdc++.h>
using namespace std;

// structure of a node of BST
struct Node {
int data;
Node* left, *right;
};

// function to create and return a node of BST
Node* getNode(int data)
{
// allocate space for the node
Node* new_node = (Node*)malloc(sizeof(Node));

// put in the data
new_node->data = data;
new_node->left = new_node->right = NULL;
}

// function to count pairs from two BSTs
// whose sum is equal to a given value x
int countPairs(Node* root1, Node* root2, int x)
{
// if either of the tree is empty
if (root1 == NULL || root2 == NULL)
return 0;

// stack 'st1' used for the inorder
// traversal of BST 1
// stack 'st2' used for the reverse
// inorder traversal of BST 2
stack<Node*> st1, st2;
Node* top1, *top2;

int count = 0;

// the loop will break when either of two
// traversals gets completed
while (1) {

// to find next node in inorder
// traversal of BST 1
while (root1 != NULL) {
st1.push(root1);
root1 = root1->left;
}

// to find next node in reverse
// inorder traversal of BST 2
while (root2 != NULL) {
st2.push(root2);
root2 = root2->right;
}

// if either gets empty then corresponding
// tree traversal is completed
if (st1.empty() || st2.empty())
break;

top1 = st1.top();
top2 = st2.top();

// if the sum of the node's is equal to 'x'
if ((top1->data + top2->data) == x) {
// increment count
count++;

// pop nodes from the respective stacks
st1.pop();
st2.pop();

// insert next possible node in the
// respective stacks
root1 = top1->right;
root2 = top2->left;
}

// move to next possible node in the
// inoder traversal of BST 1
else if ((top1->data + top2->data) < x) {
st1.pop();
root1 = top1->right;
}

// move to next possible node in the
// reverse inoder traversal of BST 2
else {
st2.pop();
root2 = top2->left;
}
}

// required count of pairs
return count;
}

// Driver program to test above
int main()
{
// formation of BST 1
Node* root1 = getNode(5); /*             5        */
root1->left = getNode(3); /*           /   \      */
root1->right = getNode(7); /*         3     7     */
root1->left->left = getNode(2); /*    / \   / \    */
root1->left->right = getNode(4); /*  2  4  6  8    */
root1->right->left = getNode(6);
root1->right->right = getNode(8);

// formation of BST 2
Node* root2 = getNode(10); /*           10         */
root2->left = getNode(6); /*           /   \        */
root2->right = getNode(15); /*        6     15      */
root2->left->left = getNode(3); /*    / \   /  \     */
root2->left->right = getNode(8); /*  3  8  11  18    */
root2->right->left = getNode(11);
root2->right->right = getNode(18);

int x = 16;
cout << "Pairs = "
<< countPairs(root1, root2, x);

return 0;
}

Java

// Java implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
import java.util.Stack;
public class GFG {

// structure of a node of BST
static class Node {
int data;
Node left, right;

// constructor
public Node(int data) {
this.data = data;
left = null;
right = null;
}
}

static Node root1;
static Node root2;
// function to count pairs from two BSTs
// whose sum is equal to a given value x
static int countPairs(Node root1, Node root2,
int x)
{
// if either of the tree is empty
if (root1 == null || root2 == null)
return 0;

// stack 'st1' used for the inorder
// traversal of BST 1
// stack 'st2' used for the reverse
// inorder traversal of BST 2
//stack<Node*> st1, st2;
Stack<Node> st1 = new Stack<>();
Stack<Node> st2 = new Stack<>();
Node top1, top2;

int count = 0;

// the loop will break when either of two
// traversals gets completed
while (true) {

// to find next node in inorder
// traversal of BST 1
while (root1 != null) {
st1.push(root1);
root1 = root1.left;
}

// to find next node in reverse
// inorder traversal of BST 2
while (root2 != null) {
st2.push(root2);
root2 = root2.right;
}

// if either gets empty then corresponding
// tree traversal is completed
if (st1.empty() || st2.empty())
break;

top1 = st1.peek();
top2 = st2.peek();

// if the sum of the node's is equal to 'x'
if ((top1.data + top2.data) == x) {
// increment count
count++;

// pop nodes from the respective stacks
st1.pop();
st2.pop();

// insert next possible node in the
// respective stacks
root1 = top1.right;
root2 = top2.left;
}

// move to next possible node in the
// inoder traversal of BST 1
else if ((top1.data + top2.data) < x) {
st1.pop();
root1 = top1.right;
}

// move to next possible node in the
// reverse inoder traversal of BST 2
else {
st2.pop();
root2 = top2.left;
}
}

// required count of pairs
return count;
}

// Driver program to test above
public static void main(String args[])
{
// formation of BST 1
root1 = new Node(5);       /*             5        */
root1.left = new Node(3); /*           /   \      */
root1.right = new Node(7); /*         3     7     */
root1.left.left = new Node(2); /*    / \   / \    */
root1.left.right = new Node(4); /*  2   4 6   8    */
root1.right.left = new Node(6);
root1.right.right = new Node(8);

// formation of BST 2
root2 = new Node(10);        /*           10         */
root2.left = new Node(6); /*           /   \        */
root2.right = new Node(15); /*        6     15      */
root2.left.left = new Node(3); /*    / \   /  \     */
root2.left.right = new Node(8); /*  3  8  11  18    */
root2.right.left = new Node(11);
root2.right.right = new Node(18);

int x = 16;
System.out.println("Pairs = "
+ countPairs(root1, root2, x));
}
}
// This code is contributed by Sumit Ghosh

Output:

Pairs = 3

Time Complexity: O(n1 + n2)
Auxiliary Space: O(n1 + n2)

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