# Find a pair with given sum in a Balanced BST

Given a Balanced Binary Search Tree and a target sum, write a function that returns true if there is a pair with sum equals to target sum, otherwise return false. Expected time complexity is O(n) and only O(Logn) extra space can be used. Any modification to Binary Search Tree is not allowed. Note that height of a Balanced BST is always O(Logn).

This problem is mainly extension of the previous post. Here we are not allowed to modify the BST.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The Brute Force Solution is to consider each pair in BST and check whether the sum equals to X. The time complexity of this solution will be O(n^2).

A Better Solution is to create an auxiliary array and store Inorder traversal of BST in the array. The array will be sorted as Inorder traversal of BST always produces sorted data. Once we have the Inorder traversal, we can pair in O(n) time (See this for details). This solution works in O(n) time, but requires O(n) auxiliary space.

```// Java code to find a pair with given sum
// in a Balanced BST
import java.util.ArrayList;

// A binary tree node
class Node {

int data;
Node left, right;

Node(int d) {
data = d;
left = right = null;
}
}

class BinarySearchTree
{

// Root of BST
Node root;

// Constructor
BinarySearchTree() {
root = null;
}

// Inorder traversal of the tree
void inorder()
{
inorderUtil(this.root);
}

// Utility function for inorder traversal of the tree
void inorderUtil(Node node)
{
if(node == null)
return;

inorderUtil(node.left);
System.out.print(node.data + " ");
inorderUtil(node.right);
}

// This method mainly calls insertRec()
void insert(int key) {
root = insertRec(root, key);
}

/* A recursive function to insert a new key in BST */
Node insertRec(Node root, int data) {

/* If the tree is empty, return a new node */
if (root == null) {
root = new Node(data);
return root;
}

/* Otherwise, recur down the tree */
if (data < root.data)
root.left = insertRec(root.left, data);
else if (data > root.data)
root.right = insertRec(root.right, data);

return root;
}

// Method that adds values of given BST into ArrayList
// and hence returns the ArrayList
ArrayList<Integer> treeToList(Node node, ArrayList<Integer>
list)
{
// Base Case
if(node == null)
return list;

treeToList(node.left, list);
treeToList(node.right, list);

return list;
}

// method that checks if there is a pair present
boolean isPairPresent(Node node, int target)
{
// This list a1 is passed as an argument
// in treeToList method
// which is later on filled by the values of BST
ArrayList<Integer>a1 = new ArrayList<>();

// a2 list contains all the values of BST
// returned by treeToList method
ArrayList<Integer> a2 = treeToList(node, a1);

int start = 0; // Starting index of a2

int end = a2.size() - 1; // Ending index of a2

while(start < end)
{

if(a2.get(start) + a2.get(end) == target) // Target Found!
{
System.out.println("Pair Found: "+a2.get(start)+
" + "+a2.get(end)+" " + "= "+ target);
return true;
}

if(a2.get(start) + a2.get(end)>target) // decrements end
{
end--;
}

if(a2.get(start) + a2.get(end)<target) // increments start
{
start++;
}
}

System.out.println("No such values are found!");
return false;
}

// Driver function
public static void main(String[] args) {
BinarySearchTree tree = new BinarySearchTree();
/*
15
/     \
10      20
/ \     /  \
8  12   16  25    */
tree.insert(15);
tree.insert(10);
tree.insert(20);
tree.insert(8);
tree.insert(12);
tree.insert(16);
tree.insert(25);

tree.isPairPresent(tree.root, 33);
}
}

// This code is contributed by Kamal Rawal
```

Output :

```Pair Found: 8 + 25 = 33
```

A space optimized solution is discussed in previous post. The idea was to first in-place convert BST to Doubly Linked List (DLL), then find pair in sorted DLL in O(n) time. This solution takes O(n) time and O(Logn) extra space, but it modifies the given BST.

The solution discussed below takes O(n) time, O(Logn) space and doesn’t modify BST. The idea is same as finding the pair in sorted array (See method 1 of this for details). We traverse BST in Normal Inorder and Reverse Inorder simultaneously. In reverse inorder, we start from the rightmost node which is the maximum value node. In normal inorder, we start from the left most node which is minimum value node. We add sum of current nodes in both traversals and compare this sum with given target sum. If the sum is same as target sum, we return true. If the sum is more than target sum, we move to next node in reverse inorder traversal, otherwise we move to next node in normal inorder traversal. If any of the traversals is finished without finding a pair, we return false. Following is C++ implementation of this approach.

```/* In a balanced binary search tree isPairPresent two element which sums to
a given value time O(n) space O(logn) */
#include <stdio.h>
#include <stdlib.h>
#define MAX_SIZE 100

// A BST node
struct node
{
int val;
struct node *left, *right;
};

// Stack type
struct Stack
{
int size;
int top;
struct node* *array;
};

// A utility function to create a stack of given size
struct Stack* createStack(int size)
{
struct Stack* stack =
(struct Stack*) malloc(sizeof(struct Stack));
stack->size = size;
stack->top = -1;
stack->array =
(struct node**) malloc(stack->size * sizeof(struct node*));
return stack;
}

// BASIC OPERATIONS OF STACK
int isFull(struct Stack* stack)
{   return stack->top - 1 == stack->size;  }

int isEmpty(struct Stack* stack)
{   return stack->top == -1;   }

void push(struct Stack* stack, struct node* node)
{
if (isFull(stack))
return;
stack->array[++stack->top] = node;
}

struct node* pop(struct Stack* stack)
{
if (isEmpty(stack))
return NULL;
return stack->array[stack->top--];
}

// Returns true if a pair with target sum exists in BST, otherwise false
bool isPairPresent(struct node *root, int target)
{
// Create two stacks. s1 is used for normal inorder traversal
// and s2 is used for reverse inorder traversal
struct Stack* s1 = createStack(MAX_SIZE);
struct Stack* s2 = createStack(MAX_SIZE);

// Note the sizes of stacks is MAX_SIZE, we can find the tree size and
// fix stack size as O(Logn) for balanced trees like AVL and Red Black
// tree. We have used MAX_SIZE to keep the code simple

// done1, val1 and curr1 are used for normal inorder traversal using s1
// done2, val2 and curr2 are used for reverse inorder traversal using s2
bool done1 = false, done2 = false;
int val1 = 0, val2 = 0;
struct node *curr1 = root, *curr2 = root;

// The loop will break when we either find a pair or one of the two
// traversals is complete
while (1)
{
// Find next node in normal Inorder traversal. See following post
// http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion/
while (done1 == false)
{
if (curr1 != NULL)
{
push(s1, curr1);
curr1 = curr1->left;
}
else
{
if (isEmpty(s1))
done1 = 1;
else
{
curr1 = pop(s1);
val1 = curr1->val;
curr1 = curr1->right;
done1 = 1;
}
}
}

// Find next node in REVERSE Inorder traversal. The only
// difference between above and below loop is, in below loop
// right subtree is traversed before left subtree
while (done2 == false)
{
if (curr2 != NULL)
{
push(s2, curr2);
curr2 = curr2->right;
}
else
{
if (isEmpty(s2))
done2 = 1;
else
{
curr2 = pop(s2);
val2 = curr2->val;
curr2 = curr2->left;
done2 = 1;
}
}
}

// If we find a pair, then print the pair and return. The first
// condition makes sure that two same values are not added
if ((val1 != val2) && (val1 + val2) == target)
{
printf("\n Pair Found: %d + %d = %d\n", val1, val2, target);
return true;
}

// If sum of current values is smaller, then move to next node in
// normal inorder traversal
else if ((val1 + val2) < target)
done1 = false;

// If sum of current values is greater, then move to next node in
// reverse inorder traversal
else if ((val1 + val2) > target)
done2 = false;

// If any of the inorder traversals is over, then there is no pair
// so return false
if (val1 >= val2)
return false;
}
}

// A utility function to create BST node
struct node * NewNode(int val)
{
struct node *tmp = (struct node *)malloc(sizeof(struct node));
tmp->val = val;
tmp->right = tmp->left =NULL;
return tmp;
}

// Driver program to test above functions
int main()
{
/*
15
/     \
10      20
/ \     /  \
8  12   16  25    */
struct node *root =  NewNode(15);
root->left = NewNode(10);
root->right = NewNode(20);
root->left->left = NewNode(8);
root->left->right = NewNode(12);
root->right->left = NewNode(16);
root->right->right = NewNode(25);

int target = 33;
if (isPairPresent(root, target) == false)
printf("\n No such values are found\n");

getchar();
return 0;
}
```

Output:

` Pair Found: 8 + 25 = 33`

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