Find a pair with given sum in BST

Given a BST and a sum, find if there is a pair with given sum.

Input : sum = 28
        Root of below tree

Output : Pair is found (16, 12)

We have discussed different approaches to find a pair with given sum in below post.Find a pair with given sum in a Balanced BST



In this post, hashing based solution is discussed. We traverse binary search tree by inorder way and insert node’s value into a set. Also check for any node, difference between given sum and node’s value in set, if it is found then pair exists otherwise it doesn’t exist.

C++

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// CPP program to find a pair with
// given sum using hashing
#include <bits/stdc++.h>
using namespace std;
  
struct Node {
    int data;
    struct Node *left, *right;
};
  
Node* NewNode(int data)
{
    Node* temp = (Node*)malloc(sizeof(Node));
    temp->data = data;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
  
Node* insert(Node* root, int key)
{
    if (root == NULL)
        return NewNode(key);
    if (key < root->data)
        root->left = insert(root->left, key);
    else
        root->right = insert(root->right, key);
    return root;
}
  
bool findpairUtil(Node* root, int sum,  unordered_set<int> &set)
{
    if (root == NULL)
        return false;
  
    if (findpairUtil(root->left, sum, set))
        return true;
  
    if (set.find(sum - root->data) != set.end()) {
        cout << "Pair is found ("
             << sum - root->data << ", "
             << root->data << ")" << endl;
        return true;
    }
    else
        set.insert(root->data);
  
    return findpairUtil(root->right, sum, set);
}
  
void findPair(Node* root, int sum)
{
    unordered_set<int> set;
    if (!findpairUtil(root, sum, set))
        cout << "Pairs do not exit" << endl;
}
  
// Driver code
int main()
{
    Node* root = NULL;
    root = insert(root, 15);
    root = insert(root, 10);
    root = insert(root, 20);
    root = insert(root, 8);
    root = insert(root, 12);
    root = insert(root, 16);
    root = insert(root, 25);
    root = insert(root, 10);
  
    int sum = 33;
    findPair(root, sum);
  
    return 0;
}

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Java














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// JAVA program to find a pair with 


// given sum using hashing 


  


import java.util.*; 


  


class GFG  


{ 


  


    static class Node 


    { 


        int data; 


        Node left, right; 


    }; 


  


    static Node NewNode(int data)  


    { 


        Node temp = new Node(); 


        temp.data = data; 


        temp.left = null; 


        temp.right = null; 


        return temp; 


    } 


  


    static Node insert(Node root, int key)  


    { 


        if (root == null) 


            return NewNode(key); 


        if (key < root.data) 


            root.left = insert(root.left, key); 


        else


            root.right = insert(root.right, key); 


        return root; 


    } 


  


    static boolean findpairUtil(Node root, int sum, 


                                HashSet<Integer> set)  


    { 


        if (root == null) 


            return false; 


  


        if (findpairUtil(root.left, sum, set)) 


            return true; 


  


        if (set.contains(sum - root.data))  


        { 


            System.out.println("Pair is found (" +  


                                (sum - root.data) + 


                                ", " + root.data + ")"); 


            return true; 


        }  


        else


            set.add(root.data); 


  


        return findpairUtil(root.right, sum, set); 


    } 


  


    static void findPair(Node root, int sum)  


    { 


        HashSet<Integer> set = new HashSet<Integer>(); 


        if (!findpairUtil(root, sum, set)) 


            System.out.print("Pairs do not exit" + "\n"); 


    } 


  


    // Driver code 


    public static void main(String[] args)  


    { 


        Node root = null; 


        root = insert(root, 15); 


        root = insert(root, 10); 


        root = insert(root, 20); 


        root = insert(root, 8); 


        root = insert(root, 12); 


        root = insert(root, 16); 


        root = insert(root, 25); 


        root = insert(root, 10); 


  


        int sum = 33; 


        findPair(root, sum); 


  


    } 


} 


  


// This code is contributed by PrinciRaj1992 



















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Python3














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# Python3 program to find a pair with  


# given sum using hashing  


import sys 


import math 


  


class Node: 


    def __init__(self,data): 


        self.data = data 


        self.left = None


        self.right = None


  


def insert(root,data): 


    if root is None: 


        return Node(data) 


    if(data < root.data): 


        root.left = insert(root.left, data) 


    if(data > root.data): 


        root.right = insert(root.right, data) 


    return root 


  


def findPairUtil(root, summ, unsorted_set): 


    if root is None: 


        return False


    if findPairUtil(root.left,summ,unsorted_set): 


        return True


    if unsorted_set and (summ-root.data) in unsorted_set: 


        print("Pair is Found ({},{})".format(summ-root.data, root.data)) 


        return True


    else: 


        unsorted_set.add(root.data) 


  


    return findPairUtil(root.right,summ, unsorted_set) 


  


def findPair(root,summ): 


    unsorted_set = set() 


    if(not findPairUtil(root,summ,unsorted_set)): 


        print("Pair do not exist!") 


  


# Driver code 


if __name__=='__main__': 


    root=None


    root = insert(root,15) 


    root = insert(root,10) 


    root = insert(root,20) 


    root = insert(root,8) 


    root = insert(root,12) 


    root = insert(root,16) 


    root = insert(root,25) 


    root = insert(root,10) 


    summ = 33


    findPair(root, summ) 


  


# This code is contributed by Vikash Kumar 37 



















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C#














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// C# program to find a pair with 


// given sum using hashing 


using System; 


using System.Collections.Generic; 


  


class GFG  


{ 


  


    class Node 


    { 


        public int data; 


        public Node left, right; 


    }; 


  


    static Node NewNode(int data)  


    { 


        Node temp = new Node(); 


        temp.data = data; 


        temp.left = null; 


        temp.right = null; 


        return temp; 


    } 


  


    static Node insert(Node root, int key)  


    { 


        if (root == null) 


            return NewNode(key); 


        if (key < root.data) 


            root.left = insert(root.left, key); 


        else


            root.right = insert(root.right, key); 


        return root; 


    } 


  


    static bool findpairUtil(Node root, int sum, 


                                HashSet<int> set)  


    { 


        if (root == null) 


            return false; 


  


        if (findpairUtil(root.left, sum, set)) 


            return true; 


  


        if (set.Contains(sum - root.data))  


        { 


            Console.WriteLine("Pair is found (" +  


                                (sum - root.data) + 


                                ", " + root.data + ")"); 


            return true; 


        }  


        else


            set.Add(root.data); 


  


        return findpairUtil(root.right, sum, set); 


    } 


  


    static void findPair(Node root, int sum)  


    { 


        HashSet<int> set = new HashSet<int>(); 


        if (!findpairUtil(root, sum, set)) 


            Console.Write("Pairs do not exit" + "\n"); 


    } 


  


    // Driver code 


    public static void Main(String[] args)  


    { 


        Node root = null; 


        root = insert(root, 15); 


        root = insert(root, 10); 


        root = insert(root, 20); 


        root = insert(root, 8); 


        root = insert(root, 12); 


        root = insert(root, 16); 


        root = insert(root, 25); 


        root = insert(root, 10); 


  


        int sum = 33; 


        findPair(root, sum); 


  


    } 


} 


  


// This code is contributed by Rajput-Ji 



















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Output:


Pair is found (8, 25)



Time Complexity is O(n).








          
          
          

          

    

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