# Pair sum in a BST

#### You are given a binary search tree and an integer ‘S’. Your task is to find all the pairs of nodes in the BST which sum to the value ‘S’. If no such pair exists, then print -1 - 1.

#### Note:

```
You can use extra space of the order of not more than O(log n).
A binary search tree (BST) is a binary tree data structure which has the following properties.
• The left subtree of a node contains only nodes with data less than the node’s data.
• The right subtree of a node contains only nodes with data greater than the node’s data.
• Both the left and right subtrees must also be binary search trees.
```

#### Note:

```
1. All the elements of the Binary Search Tree are unique.
2. You can’t use the same node value/element of BST twice.
```

#### Input Format:

```
The first line of input contains an integer ‘T’, which denotes the number of test cases. Then each test case follows.
The first line of every test case contains elements of the Binary Search Tree in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.
The second line of every test case contains a single integer ‘K’ which denotes the target value.
```

#### For example:

```
The input for the tree depicted in the below image would be :
```

```
20 10 35 5 15 30 42 -1 13 -1 -1 -1 -1 -1 -1 -1
```

#### Explanation :

```
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null (-1).
```

#### Note:

```
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
```

#### Output Format:

```
Output contains each pair of integers (pair elements separated by space). The order of different pairs, to be printed, should be in sorted order.
```

#### Note:

```
You don’t need to print anything, it has already been taken care of. Just implement the given function.
```

#### Constraints:

```
1 <= T <= 5
1 <= N <= 10 ^ 3
0 <= DATA <= 10 ^ 9
1 <= S <= 10 ^ 9
Where ‘DATA’ denotes the value of each node in the given tree.
Time limit: 1 sec
```

** **The idea is to convert this problem to sorted array problem by storing the inorder traversal of BST in an array then finding the nodes having sum equal to ‘S’.

**Approach :**

- First make a function ‘inorder’ passing the root and an array ‘arr’ to store the inorder traversal of BST as its arguments.
- Inside ‘inorder()’ :
- Store the inorder traversal of BST in ‘arr’.

- Now, in the given function, call the inorder’ function to store the inorder traversal of BST in ‘arr’ and initialize a 2D array ‘ans’ to store the pair of nodes.
- Initialise an iterator pointer ‘i’ with 0 and ‘j’ to n - 1, where n is the size of ‘arr’.
- Iterate until i < j:
- If arr[i] + arr[j] is equal to ‘S’ then store the value of nodes in the ‘ans’ and then increment ‘i’ and decrement ‘j’.
- Else if arr[i] + arr[j] > S then decrement ‘j’.
- Else increment ‘i’.

- Return ‘ans’ to the given function.

The idea is to 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 the minimum value node. We add the sum of current nodes in both traversals and compare this sum with the given target sum. If the sum is the same as the target sum, we will print the nodes. If the sum is more than the target sum, we move to the next node in reverse inorder traversal, otherwise we move to the next node in normal inorder traversal.

**Approach :**

- First check if the current node is null as a base condition, then return 0 as there are no nodes in the tree.
- Declare a 2D array, say ‘ans’ to store the pairs of nodes.
- Count the total number of nodes in the BST using function ‘countNodes’ passing the root as the argument and storing the count in a variable, say, ‘totalCount’.
- Initialize a variable, say ‘count’ to store the current count of nodes.
- Now create stacks namely ‘inorder’ and ‘revInorder’ for inorder and reverse inorder traversal respectively.
- Store the ‘root’ in a variable, say ‘currentNode’.
- Now make an iteration until ‘currentNode’ does not become null. This iteration is to store the left nodes of BST in ‘inorder’.
- After coming out of the iteration, make ‘currentNode’ again pointing to root.
- Now make an iteration until ‘currentNode’ does not become null. This iteration stores the right nodes of BST in ‘revInorder’.
- Now execute a while loop with the condition ‘count’ is less than ‘totalCount’ - 1.
- Store the node on the top of ‘inorder’ in a variable, say ‘inorderTop’.
- Store the node on the top of ‘revInorder’ in a variable, say ‘revInorderTop’.
- If ‘inorderTop’s’ data + ‘revInorderTop’s’ data is equal to the given value ‘S’.
- Store ‘inorderTop’s’ data and ‘revInorderTop’s’ data in ‘ans’.
- Now store the data of ‘inorderTop’ in a variable, say ‘top’ and pop the node from ‘inorder’ and increment the ‘count’.
- Now check if ‘top’s’ right node is present then make top = top -> right.
- And make an iteration until ‘top’ is not null.
- Push ‘top’ in ‘inorder’ and make top = top -> left.

- Now store the ‘inorderTop’s’ data in ‘top’ and pop the node from ‘revInorder’ and increment ‘count’.
- Now check if ‘top’s’ left node is present then make top = top -> left.
- And make an iteration until ‘top’ is not null.
- Push ‘top’ in ‘revInorder’ and make top = top -> right.

- Else if data of ‘inorderTop’ + data of ‘revInorderTop’ is greater than given value ‘S’.
- Store the data of ‘revInorderTop’ data in a variable, say ‘top’ and pop the node from ‘revInorder’ and increment the ‘count’.
- Now check if ‘top’s’ left node is present then make top = top -> left.
- And make an iteration until ‘top’ is not null.
- Push ‘top’ in ‘revInorder’ and make top = top -> right.

- Else if data of ‘inorderTop’ data + data of ‘revInorderTop’ data is less than given value ‘S’ do the similar steps for ‘inorder’.

- Return ‘ans’ to the function.