# 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

 `// 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 treeToList(Node node, ArrayList ` `                                                 ``list) ` `    ``{ ` `        ``// Base Case ` `        ``if` `(node == ``null``) ` `            ``return` `list; ` ` `  `        ``treeToList(node.left, list); ` `        ``list.add(node.data); ` `        ``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 a1 = ``new` `ArrayList<>(); ` ` `  `        ``// a2 list contains all the values of BST ` `        ``// returned by treeToList method ` `        ``ArrayList 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 `

## C#

 `// C# code to find a pair with given sum ` `// in a Balanced BST ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `// A binary tree node ` `public` `class` `Node { ` ` `  `    ``public` `int` `data; ` `    ``public` `Node left, right; ` ` `  `    ``public` `Node(``int` `d) ` `    ``{ ` `        ``data = d; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `public` `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); ` `        ``Console.Write(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 ` `    ``List<``int``> treeToList(Node node, List<``int``> list) ` `    ``{ ` `        ``// Base Case ` `        ``if` `(node == ``null``) ` `            ``return` `list; ` ` `  `        ``treeToList(node.left, list); ` `        ``list.Add(node.data); ` `        ``treeToList(node.right, list); ` ` `  `        ``return` `list; ` `    ``} ` ` `  `    ``// method that checks if there is a pair present ` `    ``bool` `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 ` `        ``List<``int``> a1 = ``new` `List<``int``>(); ` ` `  `        ``// a2 list contains all the values of BST ` `        ``// returned by treeToList method ` `        ``List<``int``> a2 = treeToList(node, a1); ` ` `  `        ``int` `start = 0; ``// Starting index of a2 ` ` `  `        ``int` `end = a2.Count - 1; ``// Ending index of a2 ` ` `  `        ``while` `(start < end) { ` ` `  `            ``if` `(a2[start] + a2[end] == target) ``// Target Found! ` `            ``{ ` `                ``Console.WriteLine(``"Pair Found: "` `+ a2[start] + ``" + "` `+ a2[end] + ``" "` `                                  ``+ ``"= "` `+ target); ` `                ``return` `true``; ` `            ``} ` ` `  `            ``if` `(a2[start] + a2[end] > target) ``// decrements end ` `            ``{ ` `                ``end--; ` `            ``} ` ` `  `            ``if` `(a2[start] + a2[end] < target) ``// increments start ` `            ``{ ` `                ``start++; ` `            ``} ` `        ``} ` ` `  `        ``Console.WriteLine(``"No such values are found!"``); ` `        ``return` `false``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``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 contributed by Rajput-Ji `

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.

## C++

 `/* In a balanced binary search tree  ` `isPairPresent two element which sums to  ` `a given value time O(n) space O(logn) */` `#include ` `using` `namespace` `std; ` `#define MAX_SIZE 100 ` ` `  `// A BST node ` `class` `node { ` `public``: ` `    ``int` `val; ` `    ``node *left, *right; ` `}; ` ` `  `// Stack type ` `class` `Stack { ` `public``: ` `    ``int` `size; ` `    ``int` `top; ` `    ``node** array; ` `}; ` ` `  `// A utility function to create a stack of given size ` `Stack* createStack(``int` `size) ` `{ ` `    ``Stack* stack = ``new` `Stack(); ` `    ``stack->size = size; ` `    ``stack->top = -1; ` `    ``stack->array = ``new` `node*[(stack->size * ``sizeof``(node*))]; ` `    ``return` `stack; ` `} ` ` `  `// BASIC OPERATIONS OF STACK ` `int` `isFull(Stack* stack) ` `{ ` `    ``return` `stack->top - 1 == stack->size; ` `} ` ` `  `int` `isEmpty(Stack* stack) ` `{ ` `    ``return` `stack->top == -1; ` `} ` ` `  `void` `push(Stack* stack, node* node) ` `{ ` `    ``if` `(isFull(stack)) ` `        ``return``; ` `    ``stack->array[++stack->top] = node; ` `} ` ` `  `node* pop(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(node* root, ``int` `target) ` `{ ` `    ``// Create two stacks. s1 is used for ` `    ``// normal inorder traversal and s2 is ` `    ``// used for reverse inorder traversal ` `    ``Stack* s1 = createStack(MAX_SIZE); ` `    ``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; ` `    ``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 ` `        ``// https:// 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) { ` `            ``cout << ``"Pair Found: "` `<< val1 << ``"+ "` `<< val2 << ``" = "` `<< target << endl; ` `            ``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 ` `node* NewNode(``int` `val) ` `{ ` `    ``node* tmp = ``new` `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 */` `    ``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``) ` `        ``cout << ``"\nNo such values are found\n"``; ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by rathbhupendra `

## C

 `/* In a balanced binary search tree isPairPresent two element which sums to ` `   ``a given value time O(n) space O(logn) */` `#include ` `#include ` `#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 ` `        ``// https:// 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`