Given a binary search tree, and an integer X, the task is to find if there exists a triplet with sum X. Print Yes or No correspondingly. Note that the three nodes may not necessarily be distinct.
Examples:
Input: X = 15 5 / \ 3 7 / \ / \ 2 4 6 8 Output: Yes {5, 5, 5} is one such triplet. {3, 5, 7}, {2, 5, 8}, {4, 5, 6} are some others.
Input: X = 16 1 \ 2 \ 3 \ 4 \ 5 Output: No
Simple Approach: A simple approach will be to convert the BST to a sorted array and then find the triplet using three-pointers. This will take O(N) extra space where N is the number of nodes present in the Binary Search Tree. We have already discussed a similar problem in this article which takes O(N) extra space.
Better approach: We will solve this problem using a space-efficient method by reducing the additional space complexity to O(H) where H is the height of BST. For that, we will use the two pointer technique on BST.
We will traverse all the nodes for the tree one by one and for each node, we will try to find a pair with a sum equal to (X – curr->data) where ‘curr’ is the current node of the BST we are traversing.
We will use a technique similar to the technique discussed in this article for finding a pair.
Algorithm: Traverse each node of BST one by one and for each node:
- Create a forward and backward iterator for BST. Let’s say the value of nodes they are pointing at are v1 and v2.
- Now at each step,
- If v1 + v2 = X, we found a pair, thus we will increase the count by 1.
- If v1 + v2 less than or equal to x, we will make forward iterator point to the next element.
- If v1 + v2 greater than x, we will make backward iterator point to the previous element.
- We will continue the above while the left iterator doesn’t point to a node with larger value than right node.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Node of the binary tree struct node {
int data;
node* left;
node* right;
node( int data)
{
this ->data = data;
left = NULL;
right = NULL;
}
}; // Function that returns true if a pair exists // in the binary search tree with sum equal to x bool existsPair(node* root, int x)
{ // Iterators for BST
stack<node *> it1, it2;
// Initializing forward iterator
node* c = root;
while (c != NULL)
it1.push(c), c = c->left;
// Initializing backward iterator
c = root;
while (c != NULL)
it2.push(c), c = c->right;
// Two pointer technique
while (it1.size() and it2.size()) {
// Variables to store values at
// it1 and it2
int v1 = it1.top()->data, v2 = it2.top()->data;
// Base case
if (v1 + v2 == x)
return 1;
if (v1 > v2)
break ;
// Moving forward pointer
if (v1 + v2 < x) {
c = it1.top()->right;
it1.pop();
while (c != NULL)
it1.push(c), c = c->left;
}
// Moving backward pointer
else {
c = it2.top()->left;
it2.pop();
while (c != NULL)
it2.push(c), c = c->right;
}
}
// Case when no pair is found
return 0;
} // Function that returns true if a triplet exists // in the binary search tree with sum equal to x bool existsTriplet(node* root, node* curr, int x)
{ // If current node is NULL
if (curr == NULL)
return 0;
// Conditions for existence of a triplet
return (existsPair(root, x - curr->data)
|| existsTriplet(root, curr->left, x)
|| existsTriplet(root, curr->right, x));
} // Driver code int main()
{ node* root = new node(5);
root->left = new node(3);
root->right = new node(7);
root->left->left = new node(2);
root->left->right = new node(4);
root->right->left = new node(6);
root->right->right = new node(8);
int x = 24;
if (existsTriplet(root, root, x))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
// Java implementation of the approach import java.io.*;
import java.util.*;
// Node of the binary tree class Node
{ int data;
Node left, right;
Node( int item)
{
data = item;
left = right = null ;
}
} class GFG
{ static Node root;
// Function that returns true if a pair exists
// in the binary search tree with sum equal to x
static boolean existsPair(Node root, int x)
{
// Iterators for BST
Stack<Node> it1 = new Stack<Node>();
Stack<Node> it2 = new Stack<Node>();
// Initializing forward iterator
Node c = root;
while (c != null )
{
it1.push(c);
c = c.left;
}
// Initializing backward iterator
c = root;
while (c != null )
{
it2.push(c);
c = c.right;
}
// Two pointer technique
while (it1.size() > 0 && it2.size() > 0 )
{
// Variables to store values at
// it1 and it2
int v1 = it1.peek().data;
int v2 = it2.peek().data;
// Base case
if (v1 + v2 == x)
{
return true ;
}
if (v1 > v2)
{
break ;
}
// Moving forward pointer
if (v1 + v2 < x)
{
c = it1.peek().right;
it1.pop();
while (c != null )
{
it1.push(c);
c = c.left;
}
}
// Moving backward pointer
else
{
c = it2.peek().left;
it2.pop();
while (c != null )
{
it2.push(c);
c = c.right;
}
}
}
// Case when no pair is found
return false ;
}
// Function that returns true if a triplet exists
// in the binary search tree with sum equal to x
static boolean existsTriplet(Node root,
Node curr, int x )
{
// If current node is NULL
if (curr == null )
{
return false ;
}
// Conditions for existence of a triplet
return (existsPair(root, x - curr.data) ||
existsTriplet(root, curr.left, x) ||
existsTriplet(root, curr.right, x));
}
// Driver code
public static void main (String[] args)
{
GFG tree = new GFG();
tree.root = new Node( 5 );
tree.root.left = new Node( 3 );
tree.root.right = new Node( 7 );
tree.root.left.left = new Node( 2 );
tree.root.left.right = new Node( 4 );
tree.root.right.left = new Node( 6 );
tree.root.right.right = new Node( 8 );
int x = 24 ;
if (existsTriplet(root, root, x))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
}
} // This code is contributed by avanitrachhadiya2155 |
# Python3 implementation of the approach class Node:
def __init__( self , x):
self .data = x
self .left = None
self .right = None
# Function that returns true if a pair exists # in the binary search tree with sum equal to x def existsPair(root, x):
# Iterators for BST
it1, it2 = [], []
# Initializing forward iterator
c = root
while (c ! = None ):
it1.append(c)
c = c.left
# Initializing backward iterator
c = root
while (c ! = None ):
it2.append(c)
c = c.right
# Two pointer technique
while ( len (it1) > 0 and len (it2) > 0 ):
# Variables to store values at
# it1 and it2
v1 = it1[ - 1 ].data
v2 = it2[ - 1 ].data
# Base case
if (v1 + v2 = = x):
return 1
if (v1 > v2):
break
# Moving forward pointer
if (v1 + v2 < x):
c = it1[ - 1 ].right
del it1[ - 1 ]
while (c ! = None ):
it1.append(c)
c = c.left
# Moving backward pointer
else :
c = it2[ - 1 ].left
del it2[ - 1 ]
while (c ! = None ):
it2.append(c)
c = c.right
# Case when no pair is found
return 0
# Function that returns true if a triplet exists # in the binary search tree with sum equal to x def existsTriplet(root, curr, x):
# If current node is NULL
if (curr = = None ):
return 0
# Conditions for existence of a triplet
return (existsPair(root, x - curr.data)
or existsTriplet(root, curr.left, x)
or existsTriplet(root, curr.right, x))
# Driver code if __name__ = = '__main__' :
root = Node( 5 )
root.left = Node( 3 )
root.right = Node( 7 )
root.left.left = Node( 2 )
root.left.right = Node( 4 )
root.right.left = Node( 6 )
root.right.right = Node( 8 )
x = 24
if (existsTriplet(root, root, x)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by mohit kumar 29 |
// C# implementation of the approach using System;
using System.Collections.Generic;
// Node of the binary tree class Node
{ public int data;
public Node left, right;
public Node( int item)
{
data = item;
left = right = null ;
}
} class GFG{
static Node root;
// Function that returns true if a pair exists // in the binary search tree with sum equal to x static bool existsPair(Node root, int x)
{ // Iterators for BST
Stack<Node> it1 = new Stack<Node>();
Stack<Node> it2 = new Stack<Node>();
// Initializing forward iterator
Node c = root;
while (c != null )
{
it1.Push(c);
c = c.left;
}
// Initializing backward iterator
c = root;
while (c != null )
{
it2.Push(c);
c = c.right;
}
// Two pointer technique
while (it1.Count > 0 && it2.Count > 0)
{
// Variables to store values at
// it1 and it2
int v1 = it1.Peek().data;
int v2 = it2.Peek().data;
// Base case
if (v1 + v2 == x)
{
return true ;
}
if (v1 > v2)
{
break ;
}
// Moving forward pointer
if (v1 + v2 < x)
{
c = it1.Peek().right;
it1.Pop();
while (c != null )
{
it1.Push(c);
c = c.left;
}
}
// Moving backward pointer
else
{
c = it2.Peek().left;
it2.Pop();
while (c != null )
{
it2.Push(c);
c = c.right;
}
}
}
// Case when no pair is found
return false ;
} // Function that returns true if a triplet exists // in the binary search tree with sum equal to x static bool existsTriplet(Node root, Node curr, int x)
{ // If current node is NULL
if (curr == null )
{
return false ;
}
// Conditions for existence of a triplet
return (existsPair(root, x - curr.data) ||
existsTriplet(root, curr.left, x) ||
existsTriplet(root, curr.right, x));
} // Driver code static public void Main()
{ GFG.root = new Node(5);
GFG.root.left = new Node(3);
GFG.root.right = new Node(7);
GFG.root.left.left = new Node(2);
GFG.root.left.right = new Node(4);
GFG.root.right.left = new Node(6);
GFG.root.right.right = new Node(8);
int x = 24;
if (existsTriplet(root, root, x))
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
} } // This code is contributed by rag2127 |
<script> // Javascript implementation of the approach // Node of the binary tree class node { constructor(data)
{
this .data = data;
this .left = this .right = null ;
}
} // Function to find a pair with given sum function existsPair(root, x)
{ // Iterators for BST
let it1 = [], it2 = [];
// Initializing forward iterator
let c = root;
while (c != null )
{
it1.push(c);
c = c.left;
}
// Initializing backward iterator
c = root;
while (c != null )
{
it2.push(c);
c = c.right;
}
// Two pointer technique
while (it1.length > 0 && it2.length > 0)
{
// Variables to store values at
// it1 and it2
let v1 = it1[it1.length - 1].data,
v2 = it2[it2.length - 1].data;
// Base case
if (v1 + v2 == x)
return true ;
if (v1 > v2)
{
break ;
}
// Moving forward pointer
if (v1 + v2 < x)
{
c = it1[it1.length - 1].right;
it1.pop();
while (c != null )
{
it1.push(c);
c = c.left;
}
}
// Moving backward pointer
else
{
c = it2[it2.length - 1].left;
it2.pop();
while (c != null )
{
it2.push(c);
c = c.right;
}
}
}
// Case when no pair is found
return false ;
} // Function that returns true if a // triplet exists in the binary // search tree with sum equal to x function existsTriplet(root, curr, x)
{ // If current node is NULL
if (curr == null )
{
return false ;
}
// Conditions for existence of a triplet
return (existsPair(root, x - curr.data) ||
existsTriplet(root, curr.left, x) ||
existsTriplet(root, curr.right, x));
} // Driver code let root = new node(5);
root.left = new node(3);
root.right = new node(7);
root.left.left = new node(2);
root.left.right = new node(4);
root.right.left = new node(6);
root.right.right = new node(8);
let x = 24; // Calling required function if (existsTriplet(root, root, x))
document.write( "Yes" );
else document.write( "No" );
// This code is contributed by unknown2108 </script> |
Yes
Time complexity: O(N2)
Space complexity: O(H), since H extra space has been taken.