# Minimum swap required to convert binary tree to binary search tree

Given the array representation of Complete Binary Tree i.e, if index i is the parent, index 2*i + 1 is the left child and index 2*i + 2 is the right child. The task is to find the minimum number of swap required to convert it into Binary Search Tree.

Examples:

```Input : arr[] = { 5, 6, 7, 8, 9, 10, 11 }
Output : 3
Binary tree of the given array: Swap 1: Swap node 8 with node 5.
Swap 2: Swap node 9 with node 10.
Swap 3: Swap node 10 with node 7. So, minimum 3 swaps are required.

Input : arr[] = { 1, 2, 3 }
Output : 1
Binary tree of the given array: After swapping node 1 with node 2. So, only 1 swap required.
```

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

The idea is to use the fact that inorder traversal of Binary Search Tree is in increasing order of their value.
So, find the inorder traversal of the Binary Tree and store it in the array and try to sort the array. The minimum number of swap required to get the array sorted will be the answer. Please refer below post to find minimum number of swaps required to get the array sorted.

Minimum number of swaps required to sort an array

Time Complexity: O(n log n).

 `// C++ program for Minimum swap required ` `// to convert binary tree to binary search tree ` `#include ` `using` `namespace` `std; ` ` `  `// Inorder Traversal of Binary Tree ` `void` `inorder(``int` `a[], std::vector<``int``> &v,  ` `                        ``int` `n, ``int` `index) ` `{ ` `    ``// if index is greater or equal to vector size ` `    ``if``(index >= n) ` `        ``return``; ` `    ``inorder(a, v, n, 2 * index + 1); ` `     `  `    ``// push elements in vector ` `    ``v.push_back(a[index]); ` `    ``inorder(a, v, n, 2 * index + 2); ` `} ` ` `  `// Function to find minimum swaps to sort an array ` `int` `minSwaps(std::vector<``int``> &v) ` `{ ` `    ``std::vector > t(v.size()); ` `    ``int` `ans = 0; ` `    ``for``(``int` `i = 0; i < v.size(); i++) ` `        ``t[i].first = v[i], t[i].second = i; ` `     `  `    ``sort(t.begin(), t.end()); ` `    ``for``(``int` `i = 0; i < t.size(); i++) ` `    ``{ ` `        ``// second element is equal to i ` `        ``if``(i == t[i].second) ` `            ``continue``; ` `        ``else` `        ``{ ` `            ``// swaping of elements ` `            ``swap(t[i].first, t[t[i].second].first); ` `            ``swap(t[i].second, t[t[i].second].second); ` `        ``} ` `         `  `        ``// Second is not equal to i ` `        ``if``(i != t[i].second) ` `            ``--i; ` `        ``ans++; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 5, 6, 7, 8, 9, 10, 11 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``std::vector<``int``> v; ` `    ``inorder(a, v, n, 0); ` `    ``cout << minSwaps(v) << endl; ` `} ` ` `  `// This code is contributed by code_freak `

Output:

`3`

Exercise: Can we extend this to normal binary tree, i.e., a binary tree represented using left and right pointers, and not necessarily complete?

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : code_freak

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