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.


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.

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
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)
    inorder(a, v, n, 2 * index + 1);
    // push elements in vector
    inorder(a, v, n, 2 * index + 2);
// Function to find minimum swaps to sort an array
int minSwaps(std::vector<int> &v)
    std::vector<pair<int,int> > 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)
            // 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)
    return ans;
// Driver code
int main()
    int a[] = { 5, 6, 7, 8, 9, 10, 11 };
    int n = sizeof(a) / sizeof(a[0]);
    std::vector<int> v;
    inorder(a, v, n, 0);
    cout << minSwaps(v) << endl;
// This code is contributed by code_freak




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 or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Improved By : code_freak

Article Tags :
Practice Tags :


Please write to us at to report any issue with the above content.