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.

Approach :

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.

Below is the implementation of the above approach:

C++

// C++ program for Minimum swap required
// to convert binary tree to binary search tree
#include<bits/stdc++.h>
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<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)
            continue;
        else
        {
            // swapping 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[0]);
    std::vector<int> v;
    inorder(a, v, n, 0);
    cout << minSwaps(v) << endl;
}

// This code is contributed by code_freak

Java

// Java program for Minimum swap required
// to convert binary tree to binary search tree
import java.util.*;

public class GFG{

    // Pair class
    static class Pair{
        int first, second;

        Pair(int a, int b){
            first = a;
            second = b;
        }
    }

    // Inorder Traversal of Binary Tree
    static void inorder(int a[], Vector<Integer> 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.add(a[index]);

        inorder(a, v, n, 2 * index + 2);
    }

    // Function returns the
    // minimum number of swaps
    // required to sort the array
    // Refer :
    // https://www.geeksforgeeks.org/minimum-number-swaps-required-sort-array/
    public static int minSwaps(Vector<Integer> arr)
    {
        int n = arr.size();

        ArrayList < Pair > arrpos = new ArrayList < Pair > ();
        for (int i = 0; i < n; i++)
             arrpos.add(new Pair(arr.get(i), i));

        // Sort the array by array element values to
        // get right position of every element as the
        // elements of second array.
        arrpos.sort(new Comparator<Pair>()
        {
            @Override
            public int compare(Pair o1, Pair o2)
            {
                return o1.first - o2.first;
            }
        });

        // To keep track of visited elements. Initialize
        // all elements as not visited or false.
        Boolean[] vis = new Boolean[n];
        Arrays.fill(vis, false);

        // Initialize result
        int ans = 0;

        // Traverse array elements
        for (int i = 0; i < n; i++)
        {
            // already swapped and corrected or
            // already present at correct pos
            if (vis[i] || arrpos.get(i).second == i)
                continue;

            // find out the number of  node in
            // this cycle and add in ans
            int cycle_size = 0;
            int j = i;
            while (!vis[j])
            {
                vis[j] = true;

                // move to next node
                j = arrpos.get(j).second;
                cycle_size++;
            }

            // Update answer by adding current cycle.
            if(cycle_size > 0)
            {
                ans += (cycle_size - 1);
            }
        }

        // Return result
        return ans;
    }

    // Driver code
    public static void main(String args[])
    {
        int a[] = { 5, 6, 7, 8, 9, 10, 11 };
        int n = a.length;

        Vector<Integer> v = new Vector<Integer>();

        inorder(a, v, n, 0);

        System.out.println(minSwaps(v));
    }
}

Python3

# Python3 program for Minimum swap required
# to convert binary tree to binary search tree

# Inorder Traversal of Binary Tree
def inorder(a, n, index):

    global v

    # If index is greater or equal to
    # vector size
    if (index >= n):
        return

    inorder(a, n, 2 * index + 1)

    # Push elements in vector
    v.append(a[index])
    inorder(a, n, 2 * index + 2)

# Function to find minimum swaps
# to sort an array
def minSwaps():

    global v
    t = [[0, 0] for i in range(len(v))]
    ans = -2

    for i in range(len(v)):
        t[i][0], t[i][1] = v[i], i

    t, i = sorted(t), 0

    while i < len(t):

        # break
        # second element is equal to i
        if (i == t[i][1]):
            i += 1
            continue
        else:

            # Swapping of elements
            t[i][0], t[t[i][1]][0] = t[t[i][1]][0], t[i][0]
            t[i][1], t[t[i][1]][1] = t[t[i][1]][1], t[i][1]

        # Second is not equal to i
        if (i == t[i][1]):
            i -= 1

        i += 1

        ans += 1

    return ans

# Driver Code
if __name__ == '__main__':

    v = []
    a = [ 5, 6, 7, 8, 9, 10, 11 ]
    n = len(a)
    inorder(a, n, 0)

    print(minSwaps())

# This code is contributed by mohit kumar 29

C#

// C# program for Minimum swap required
// to convert binary tree to binary search tree
using System;
using System.Collections.Generic;
using System.Linq;

class GFG {

    // Pair class
    public class Pair {
        public int first, second;

        public Pair(int a, int b)
        {
            first = a;
            second = b;
        }
    }

    // Inorder Traversal of Binary Tree
    public static void inorder(int[] a, List<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.Add(a[index]);

        inorder(a, v, n, 2 * index + 2);
    }

    // Function returns the
    // minimum number of swaps
    // required to sort the array
    // Refer :
    // https://www.geeksforgeeks.org/minimum-number-swaps-required-sort-array/

    public static int minSwaps(List<int> arr)
    {

        int n = arr.Count();

        List<Pair> arrpos = new List<Pair>();

        for (int i = 0; i < n; i++)
            arrpos.Add(new Pair(arr[i], i));

        // Sort the array by array element values to
        // get right position of every element as the
        // elements of second array.
        arrpos.Sort((x, y) => x.first - y.first);

        // To keep track of visited elements. Initialize
        // all elements as not visited or false.
        bool[] vis = new bool[n];

        for (int i = 0; i < n; i++)
            vis[i] = false;

        // Initialize result
        int ans = 0;

        // Traverse array elements
        for (int i = 0; i < n; i++) {
            // already swapped and corrected or
            // already present at correct pos
            if (vis[i] || arrpos[i].first == i)
                continue;

            // find out the number of  node in
            // this cycle and add in ans
            int cycle_size = 0;
            int j = i;
            while (!vis[j]) {
                vis[j] = true;

                // move to next node
                j = arrpos[j].second;
                cycle_size++;
            }
            // Update answer by adding current cycle.
            if (cycle_size > 0)
                ans += (cycle_size - 1);
        }
        // Return result
        return ans;
    }

    // Driver code
    public static void Main(string[] args)
    {
        int[] a = { 5, 6, 7, 8, 9, 10, 11 };
        int n = a.Length;

        List<int> v = new List<int>();

        inorder(a, v, n, 0);

        Console.WriteLine(minSwaps(v));
    }
}

// This Code is contributed by Tapesh(tapeshdua420)

Javascript

<script>
// Javascript program for Minimum swap required
// to convert binary tree to binary search tree

// Inorder Traversal of Binary Tree
function inorder(a, n, index)
{
    // If index is greater or equal to
    // vector size
    if (index >= n)
        return

    inorder(a, n, 2 * index + 1)

    // Push elements in vector
    v.push(a[index])
    inorder(a, n, 2 * index + 2)
}

// Function to find minimum swaps to sort an array
function minSwaps()
{
    let t=new Array(v.length);
    let ans = -2

    for(let i=0;i<v.length;i++)
    {
        t[i]=new Array(2);
        for(let j=0;j<2;j++)
        {
            t[i][j]=0;
        }
    }

    for(let i=0;i<v.length;i++)
    {
        t[i][0]=v[i];
        t[i][1]=i;
    }

    t.sort(function(a,b){return a[0] - b[0];});
    let i=0;

    while(i<t.length)
    {

        // break
        // second element is equal to i
        if (i == t[i][1])
        {    i += 1;
            continue;
        }
        else{

            // Swapping of elements
            t[i][0], t[t[i][1]][0] = t[t[i][1]][0], t[i][0];
            t[i][1], t[t[i][1]][1] = t[t[i][1]][1], t[i][1];
         }
        // Second is not equal to i
        if (i == t[i][1])
            i -= 1;

        i += 1;

        ans += 1;
    }


     return ans;

}
// Driver code
let v=[];
let a=[ 5, 6, 7, 8, 9, 10, 11];
let n=a.length;
inorder(a, n, 0);
document.write(minSwaps());


// This code is contributed by patel2127
</script>

Output

3

Time Complexity: O(n*logn)
Auxiliary Space: O(n) because it is using extra space for array 

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

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