# Find the minimum absolute difference in two different BST’s

Given 2 Binary Search Trees, select one node from each tree such that their absolute difference is minimum possible. Assume each BST has at-least one node.

Examples:

```Input : N1 = 7, N2 = 2

BST1 :
5
/   \
3     7
/ \   / \
2   4 6   8

BST2 :
11
\
13

Output : 3
8 is largest number in the first BST
and 11 is smallest in the second.
Thus, the final answer will be 11-8 = 3

Input : N1 = 4, N2 = 2
BST1 :
3
/   \
2     4
\
14

BST2 :
7
\
13

Output : 1
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The idea is to use the two-pointer technique and iterating the pointers using the following steps.

1. Create forward iterators for both the BST’s. Let’s say that the value of nodes they are pointing at are v1 and v2 respectively.
2. Now at each step:
• Update final ans as min(ans, abs(v1-v2)) .
• If v1 < v2, move iterator of first BST else move the iterator of the second BST.
3. Repeat above steps till both the BST’s are pointing to a valid nodes.

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Node of Binary tree ` `struct` `node { ` `    ``int` `data; ` `    ``node* left; ` `    ``node* right; ` `    ``node(``int` `data) ` `    ``{ ` `        ``this``->data = data; ` `        ``left = NULL; ` `        ``right = NULL; ` `    ``} ` `}; ` ` `  `// Function to iterate to the ` `// next element of the BST ` `void` `next(stack& it) ` `{ ` ` `  `    ``node* curr = it.top()->right; ` `    ``it.pop(); ` `    ``while` `(curr != NULL) ` `        ``it.push(curr), curr = curr->left; ` `} ` ` `  `// Function to find minimum difference ` `int` `minDiff(node* root1, node* root2) ` `{ ` ` `  `    ``// Iterator for two Binary Search Trees ` `    ``stack it1, it2; ` ` `  `    ``// Initializing first iterator ` `    ``node* curr = root1; ` `    ``while` `(curr != NULL) ` `        ``it1.push(curr), curr = curr->left; ` ` `  `    ``// Initializing second iterator ` `    ``curr = root2; ` `    ``while` `(curr != NULL) ` `        ``it2.push(curr), curr = curr->left; ` ` `  `    ``// Variable to store final answer ` `    ``int` `ans = INT_MAX; ` ` `  `    ``// Two pointer technique ` `    ``while` `(it1.size() and it2.size()) { ` ` `  `        ``// value it1 and it2 are pointing to ` `        ``int` `v1 = it1.top()->data; ` `        ``int` `v2 = it2.top()->data; ` ` `  `        ``// Updating final answer ` `        ``ans = min(``abs``(v1 - v2), ans); ` ` `  `        ``// Case when v1 < v2 ` `        ``if` `(v1 < v2) ` `            ``next(it1); ` `        ``else` `            ``next(it2); ` `    ``} ` ` `  `    ``// Return ans ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// BST-1 ` ` `  `    ``/*    5  ` `        ``/   \  ` `       ``3     7  ` `      ``/ \   / \  ` `     ``2   4 6   8 */` `    ``node* root2 = ``new` `node(5); ` `    ``root2->left = ``new` `node(3); ` `    ``root2->right = ``new` `node(7); ` `    ``root2->left->left = ``new` `node(2); ` `    ``root2->left->right = ``new` `node(4); ` `    ``root2->right->left = ``new` `node(6); ` `    ``root2->right->right = ``new` `node(8); ` ` `  `    ``// BST-2 ` ` `  `    ``/*  11 ` `         ``\ ` `          ``15 ` `    ``*/` `    ``node* root1 = ``new` `node(11); ` `    ``root1->right = ``new` `node(15); ` ` `  `    ``cout << minDiff(root1, root2); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Node of Binary tree ` `static` `class` `node  ` `{ ` `    ``int` `data; ` `    ``node left; ` `    ``node right; ` `    ``node(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `        ``left = ``null``; ` `        ``right = ``null``; ` `    ``} ` `}; ` ` `  `// Function to iterate to the ` `// next element of the BST ` `static` `void` `next(Stack it) ` `{ ` `    ``node curr = it.peek().right; ` `    ``it.pop(); ` `    ``while` `(curr != ``null``) ` `    ``{ ` `        ``it.push(curr); ` `        ``curr = curr.left; ` `    ``} ` `} ` ` `  `// Function to find minimum difference ` `static` `int` `minDiff(node root1, node root2) ` `{ ` ` `  `    ``// Iterator for two Binary Search Trees ` `    ``Stack it1 = ``new` `Stack(); ` `    ``Stack it2 = ``new` `Stack(); ` ` `  `    ``// Initializing first iterator ` `    ``node curr = root1; ` `    ``while` `(curr != ``null``) ` `    ``{ ` `        ``it1.push(curr); ` `        ``curr = curr.left; ` `    ``} ` ` `  `    ``// Initializing second iterator ` `    ``curr = root2; ` `    ``while` `(curr != ``null``) ` `    ``{ ` `        ``it2.push(curr); ` `        ``curr = curr.left; ` `    ``} ` ` `  `    ``// Variable to store final answer ` `    ``int` `ans = Integer.MAX_VALUE; ` ` `  `    ``// Two pointer technique ` `    ``while` `(it1.size() > ``0` `&& it2.size() > ``0``)  ` `    ``{ ` ` `  `        ``// value it1 and it2 are pointing to ` `        ``int` `v1 = it1.peek().data; ` `        ``int` `v2 = it2.peek().data; ` ` `  `        ``// Updating final answer ` `        ``ans = Math.min(Math.abs(v1 - v2), ans); ` ` `  `        ``// Case when v1 < v2 ` `        ``if` `(v1 < v2) ` `            ``next(it1); ` `        ``else` `            ``next(it2); ` `    ``} ` ` `  `    ``// Return ans ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``// BST-1 ` ` `  `    ``/* 5  ` `        ``/ \  ` `    ``3     7  ` `    ``/ \ / \  ` `    ``2 4 6 8 */` `    ``node root2 = ``new` `node(``5``); ` `    ``root2.left = ``new` `node(``3``); ` `    ``root2.right = ``new` `node(``7``); ` `    ``root2.left.left = ``new` `node(``2``); ` `    ``root2.left.right = ``new` `node(``4``); ` `    ``root2.right.left = ``new` `node(``6``); ` `    ``root2.right.right = ``new` `node(``8``); ` ` `  `    ``// BST-2 ` ` `  `    ``/* 11 ` `        ``\ ` `        ``15 ` `    ``*/` `    ``node root1 = ``new` `node(``11``); ` `    ``root1.right = ``new` `node(``15``); ` ` `  `    ``System.out.println(minDiff(root1, root2)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` ` `  `// Node of Binary tree ` `class` `node  ` `{ ` `    ``public` `int` `data; ` `    ``public` `node left; ` `    ``public` `node right; ` `    ``public` `node(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `        ``left = ``null``; ` `        ``right = ``null``; ` `    ``} ` `}; ` ` `  `// Function to iterate to the ` `// next element of the BST ` `static` `void` `next(Stack it) ` `{ ` `    ``node curr = it.Peek().right; ` `    ``it.Pop(); ` `    ``while` `(curr != ``null``) ` `    ``{ ` `        ``it.Push(curr); ` `        ``curr = curr.left; ` `    ``} ` `} ` ` `  `// Function to find minimum difference ` `static` `int` `minDiff(node root1, node root2) ` `{ ` ` `  `    ``// Iterator for two Binary Search Trees ` `    ``Stack it1 = ``new` `Stack(); ` `    ``Stack it2 = ``new` `Stack(); ` ` `  `    ``// Initializing first iterator ` `    ``node curr = root1; ` `    ``while` `(curr != ``null``) ` `    ``{ ` `        ``it1.Push(curr); ` `        ``curr = curr.left; ` `    ``} ` ` `  `    ``// Initializing second iterator ` `    ``curr = root2; ` `    ``while` `(curr != ``null``) ` `    ``{ ` `        ``it2.Push(curr); ` `        ``curr = curr.left; ` `    ``} ` ` `  `    ``// Variable to store readonly answer ` `    ``int` `ans = ``int``.MaxValue; ` ` `  `    ``// Two pointer technique ` `    ``while` `(it1.Count > 0 && it2.Count > 0)  ` `    ``{ ` ` `  `        ``// value it1 and it2 are pointing to ` `        ``int` `v1 = it1.Peek().data; ` `        ``int` `v2 = it2.Peek().data; ` ` `  `        ``// Updating readonly answer ` `        ``ans = Math.Min(Math.Abs(v1 - v2), ans); ` ` `  `        ``// Case when v1 < v2 ` `        ``if` `(v1 < v2) ` `            ``next(it1); ` `        ``else` `            ``next(it2); ` `    ``} ` ` `  `    ``// Return ans ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``// BST-1 ` ` `  `    ``/* 5  ` `        ``/ \  ` `    ``3     7  ` `    ``/ \ / \  ` `    ``2 4 6 8 */` `    ``node root2 = ``new` `node(5); ` `    ``root2.left = ``new` `node(3); ` `    ``root2.right = ``new` `node(7); ` `    ``root2.left.left = ``new` `node(2); ` `    ``root2.left.right = ``new` `node(4); ` `    ``root2.right.left = ``new` `node(6); ` `    ``root2.right.right = ``new` `node(8); ` ` `  `    ``// BST-2 ` ` `  `    ``/* 11 ` `        ``\ ` `        ``15 ` `    ``*/` `    ``node root1 = ``new` `node(11); ` `    ``root1.right = ``new` `node(15); ` ` `  `    ``Console.WriteLine(minDiff(root1, root2)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```3
```

Time complexity: O(N1 + N2) where N1 and N2 are the bumber of nodes of first and second BST respectively.
Space Complexity: O(H1 + H2) where H1 and H2 are the heights of first and second BST respectively.

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Improved By : 29AjayKumar, Rajput-Ji