Given two Binary Search Trees (BST) and a given sum. The task is to find pairs with given sum such that each pair elements must lie in different BST.
Examples:
Input : sum = 10
8 5
/ \ / \
3 10 2 18
/ \ \ / \
1 6 14 1 3
/ \ / \
5 7 13 4
Output : (5,5), (6,4), (7,3), (8,2)
In above pairs first element lies in first
BST and second element lies in second BST
A simple solution for this problem is to store Inorder traversal of one tree in auxiliary array then pick element one by one from the array and find its pair in other tree for given sum. Time complexity for this approach will be O(n2) if total nodes in both the trees are equal.
An efficient solution for this solution is to store Inorder traversals of both BSTs in two different auxiliary arrays vect1[] and vect2[]. Now we follow method1 of this article. Since Inorder traversal of BST is always gives sorted sequence, we don not need to sort our arrays.
- Take iterator left and points it to the left corner vect1[].
- Take iterator right and points it to the right corner vect2[].
- Now if vect11[left] + vect2[right] < sum then move left iterator in vect1[] in forward direction i.e; left++.
- Now if vect1[left] + vect2[right] > sum then move right iterator in vect[] in backward direction i.e; right–.
Below is implementation of above idea.
C++
// C++ program to find pairs with given sum such // that one element of pair exists in one BST and // other in other BST. #include<bits/stdc++.h> using namespace std; // A binary Tree node struct Node { int data; struct Node *left, *right; }; // A utility function to create a new BST node // with key as given num struct Node* newNode(int num) { struct Node* temp = new Node; temp->data = num; temp->left = temp->right = NULL; return temp; } // A utility function to insert a given key to BST Node* insert(Node* root, int key) { if (root == NULL) return newNode(key); if (root->data > key) root->left = insert(root->left, key); else root->right = insert(root->right, key); return root; } // store storeInorder traversal in auxiliary array void storeInorder(Node *ptr, vector<int> &vect) { if (ptr==NULL) return; storeInorder(ptr->left, vect); vect.push_back(ptr->data); storeInorder(ptr->right, vect); } // Function to find pair for given sum in different bst // vect1[] --> stores storeInorder traversal of first bst // vect2[] --> stores storeInorder traversal of second bst void pairSumUtil(vector<int> &vect1, vector<int> &vect2, int sum) { // Initialize two indexes to two different corners // of two vectors. int left = 0; int right = vect2.size() - 1; // find pair by moving two corners. while (left < vect1.size() && right >= 0) { // If we found a pair if (vect1[left] + vect2[right] == sum) { cout << "(" << vect1[left] << ", " << vect2[right] << "), "; left++; right--; } // If sum is more, move to higher value in // first vector. else if (vect1[left] + vect2[right] < sum) left++; // If sum is less, move to lower value in // second vector. else right--; } } // Prints all pairs with given "sum" such that one // element of pair is in tree with root1 and other // node is in tree with root2. void pairSum(Node *root1, Node *root2, int sum) { // Store inorder traversals of two BSTs in two // vectors. vector<int> vect1, vect2; storeInorder(root1, vect1); storeInorder(root2, vect2); // Now the problem reduces to finding a pair // with given sum such that one element is in // vect1 and other is in vect2. pairSumUtil(vect1, vect2, sum); } // Driver program to run the case int main() { // first BST struct Node* root1 = NULL; root1 = insert(root1, 8); root1 = insert(root1, 10); root1 = insert(root1, 3); root1 = insert(root1, 6); root1 = insert(root1, 1); root1 = insert(root1, 5); root1 = insert(root1, 7); root1 = insert(root1, 14); root1 = insert(root1, 13); // second BST struct Node* root2 = NULL; root2 = insert(root2, 5); root2 = insert(root2, 18); root2 = insert(root2, 2); root2 = insert(root2, 1); root2 = insert(root2, 3); root2 = insert(root2, 4); int sum = 10; pairSum(root1, root2, sum); return 0; } |
Java
// Java program to find pairs with given sum such // that one element of pair exists in one BST and // other in other BST. import java.util.*; class solution { // A binary Tree node static class Node { int data; Node left, right; }; // A utility function to create a new BST node // with key as given num static Node newNode(int num) { Node temp = new Node(); temp.data = num; temp.left = temp.right = null; return temp; } // A utility function to insert a given key to BST static Node insert(Node root, int key) { if (root == null) return newNode(key); if (root.data > key) root.left = insert(root.left, key); else root.right = insert(root.right, key); return root; } // store storeInorder traversal in auxiliary array static void storeInorder(Node ptr, Vector<Integer> vect) { if (ptr==null) return; storeInorder(ptr.left, vect); vect.add(ptr.data); storeInorder(ptr.right, vect); } // Function to find pair for given sum in different bst // vect1.get() -. stores storeInorder traversal of first bst // vect2.get() -. stores storeInorder traversal of second bst static void pairSumUtil(Vector<Integer> vect1, Vector<Integer> vect2, int sum) { // Initialize two indexes to two different corners // of two Vectors. int left = 0; int right = vect2.size() - 1; // find pair by moving two corners. while (left < vect1.size() && right >= 0) { // If we found a pair if (vect1.get(left) + vect2.get(right) == sum) { System.out.print( "(" +vect1.get(left) + ", "+ vect2.get(right) + "), "); left++; right--; } // If sum is more, move to higher value in // first Vector. else if (vect1.get(left) + vect2.get(right) < sum) left++; // If sum is less, move to lower value in // second Vector. else right--; } } // Prints all pairs with given "sum" such that one // element of pair is in tree with root1 and other // node is in tree with root2. static void pairSum(Node root1, Node root2, int sum) { // Store inorder traversals of two BSTs in two // Vectors. Vector<Integer> vect1= new Vector<Integer>(), vect2= new Vector<Integer>(); storeInorder(root1, vect1); storeInorder(root2, vect2); // Now the problem reduces to finding a pair // with given sum such that one element is in // vect1 and other is in vect2. pairSumUtil(vect1, vect2, sum); } // Driver program to run the case public static void main(String args[]) { // first BST Node root1 = null; root1 = insert(root1, 8); root1 = insert(root1, 10); root1 = insert(root1, 3); root1 = insert(root1, 6); root1 = insert(root1, 1); root1 = insert(root1, 5); root1 = insert(root1, 7); root1 = insert(root1, 14); root1 = insert(root1, 13); // second BST Node root2 = null; root2 = insert(root2, 5); root2 = insert(root2, 18); root2 = insert(root2, 2); root2 = insert(root2, 1); root2 = insert(root2, 3); root2 = insert(root2, 4); int sum = 10; pairSum(root1, root2, sum); } } //contributed by Arnab Kundu |
Python3
# Python3 program to find pairs with given # sum such that one element of pair exists # in one BST and other in other BST. # A utility function to create a new # BST node with key as given num class newNode: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # A utility function to insert a # given key to BST def insert(root, key): if root == None: return newNode(key) if root.data > key: root.left = insert(root.left, key) else: root.right = insert(root.right, key) return root # store storeInorder traversal in # auxiliary array def storeInorder(ptr, vect): if ptr == None: return storeInorder(ptr.left, vect) vect.append(ptr.data) storeInorder(ptr.right, vect) # Function to find pair for given # sum in different bst # vect1[] --> stores storeInorder traversal # of first bst # vect2[] --> stores storeInorder traversal # of second bst def pairSumUtil(vect1, vect2, Sum): # Initialize two indexes to two # different corners of two lists. left = 0 right = len(vect2) - 1 # find pair by moving two corners. while left < len(vect1) and right >= 0: # If we found a pair if vect1[left] + vect2[right] == Sum: print("(", vect1[left], ",", vect2[right], "),", end = " ") left += 1 right -= 1 # If sum is more, move to higher # value in first lists. else if vect1[left] + vect2[right] < Sum: left += 1 # If sum is less, move to lower # value in second lists. else: right -= 1 # Prints all pairs with given "sum" such that one # element of pair is in tree with root1 and other # node is in tree with root2. def pairSum(root1, root2, Sum): # Store inorder traversals of # two BSTs in two lists. vect1 = [] vect2 = [] storeInorder(root1, vect1) storeInorder(root2, vect2) # Now the problem reduces to finding a # pair with given sum such that one # element is in vect1 and other is in vect2. pairSumUtil(vect1, vect2, Sum) # Driver Code if __name__ == '__main__': # first BST root1 = None root1 = insert(root1, 8) root1 = insert(root1, 10) root1 = insert(root1, 3) root1 = insert(root1, 6) root1 = insert(root1, 1) root1 = insert(root1, 5) root1 = insert(root1, 7) root1 = insert(root1, 14) root1 = insert(root1, 13) # second BST root2 = None root2 = insert(root2, 5) root2 = insert(root2, 18) root2 = insert(root2, 2) root2 = insert(root2, 1) root2 = insert(root2, 3) root2 = insert(root2, 4) Sum = 10 pairSum(root1, root2, Sum) # This code is contributed by PranchalK |
C#
using System; using System.Collections.Generic; // C# program to find pairs with given sum such // that one element of pair exists in one BST and // other in other BST. public class solution { // A binary Tree node public class Node { public int data; public Node left, right; } // A utility function to create a new BST node // with key as given num public static Node newNode(int num) { Node temp = new Node(); temp.data = num; temp.left = temp.right = null; return temp; } // A utility function to insert a given key to BST public static Node insert(Node root, int key) { if (root == null) { return newNode(key); } if (root.data > key) { root.left = insert(root.left, key); } else { root.right = insert(root.right, key); } return root; } // store storeInorder traversal in auxiliary array public static void storeInorder(Node ptr, List<int> vect) { if (ptr == null) { return; } storeInorder(ptr.left, vect); vect.Add(ptr.data); storeInorder(ptr.right, vect); } // Function to find pair for given sum in different bst // vect1.get() -. stores storeInorder traversal of first bst // vect2.get() -. stores storeInorder traversal of second bst public static void pairSumUtil(List<int> vect1, List<int> vect2, int sum) { // Initialize two indexes to two different corners // of two Vectors. int left = 0; int right = vect2.Count - 1; // find pair by moving two corners. while (left < vect1.Count && right >= 0) { // If we found a pair if (vect1[left] + vect2[right] == sum) { Console.Write("(" + vect1[left] + ", " + vect2[right] + "), "); left++; right--; } // If sum is more, move to higher value in // first Vector. else if (vect1[left] + vect2[right] < sum) { left++; } // If sum is less, move to lower value in // second Vector. else { right--; } } } // Prints all pairs with given "sum" such that one // element of pair is in tree with root1 and other // node is in tree with root2. public static void pairSum(Node root1, Node root2, int sum) { // Store inorder traversals of two BSTs in two // Vectors. List<int> vect1 = new List<int>(), vect2 = new List<int>(); storeInorder(root1, vect1); storeInorder(root2, vect2); // Now the problem reduces to finding a pair // with given sum such that one element is in // vect1 and other is in vect2. pairSumUtil(vect1, vect2, sum); } // Driver program to run the case public static void Main(string[] args) { // first BST Node root1 = null; root1 = insert(root1, 8); root1 = insert(root1, 10); root1 = insert(root1, 3); root1 = insert(root1, 6); root1 = insert(root1, 1); root1 = insert(root1, 5); root1 = insert(root1, 7); root1 = insert(root1, 14); root1 = insert(root1, 13); // second BST Node root2 = null; root2 = insert(root2, 5); root2 = insert(root2, 18); root2 = insert(root2, 2); root2 = insert(root2, 1); root2 = insert(root2, 3); root2 = insert(root2, 4); int sum = 10; pairSum(root1, root2, sum); } } // This code is contributed by Shrikant13 |
Javascript
<script> // JavaScript program to find pairs with given sum such // that one element of pair exists in one BST and // other in other BST. // A binary Tree node class Node { constructor() { this.data = 0; this.left = null; this.right = null; } } // A utility function to create a new BST node // with key as given num function newNode(num) { var temp = new Node(); temp.data = num; temp.left = temp.right = null; return temp; } // A utility function to insert a given key to BST function insert(root, key) { if (root == null) { return newNode(key); } if (root.data > key) { root.left = insert(root.left, key); } else { root.right = insert(root.right, key); } return root; } // store storeInorder traversal in auxiliary array function storeInorder(ptr, vect) { if (ptr == null) { return; } storeInorder(ptr.left, vect); vect.push(ptr.data); storeInorder(ptr.right, vect); } // Function to find pair for given sum in different bst // vect1.get() -. stores storeInorder traversal of first bst // vect2.get() -. stores storeInorder traversal of second bst function pairSumUtil(vect1, vect2, sum) { // Initialize two indexes to two different corners // of two Vectors. var left = 0; var right = vect2.length - 1; // find pair by moving two corners. while (left < vect1.length && right >= 0) { // If we found a pair if (vect1[left] + vect2[right] == sum) { document.write("(" + vect1[left] + ", " + vect2[right] + "), "); left++; right--; } // If sum is more, move to higher value in // first Vector. else if (vect1[left] + vect2[right] < sum) { left++; } // If sum is less, move to lower value in // second Vector. else { right--; } } } // Prints all pairs with given "sum" such that one // element of pair is in tree with root1 and other // node is in tree with root2. function pairSum(root1, root2, sum) { // Store inorder traversals of two BSTs in two // Vectors. var vect1 = [], vect2 = []; storeInorder(root1, vect1); storeInorder(root2, vect2); // Now the problem reduces to finding a pair // with given sum such that one element is in // vect1 and other is in vect2. pairSumUtil(vect1, vect2, sum); } // Driver program to run the case // first BST var root1 = null; root1 = insert(root1, 8); root1 = insert(root1, 10); root1 = insert(root1, 3); root1 = insert(root1, 6); root1 = insert(root1, 1); root1 = insert(root1, 5); root1 = insert(root1, 7); root1 = insert(root1, 14); root1 = insert(root1, 13); // second BST var root2 = null; root2 = insert(root2, 5); root2 = insert(root2, 18); root2 = insert(root2, 2); root2 = insert(root2, 1); root2 = insert(root2, 3); root2 = insert(root2, 4); var sum = 10; pairSum(root1, root2, sum); </script> |
Output
(5, 5), (6, 4), (7, 3), (8, 2),
Time complexity : O(n)
Auxiliary space : O(n)
We have another space optimized approach to solve this problem. The idea is to convert bst into doubly linked list and apply above method for doubly linked list.See this article.
Time complexity : O(n)
Auxiliary Space : O(1)
This article is contributed by Shashank Mishra . If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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