Minimum swap required to convert binary tree to binary search tree

07 Jul, 2022
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.
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).
Implementation:
C++
Python3
Javascript
Java
// 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 Treevoid 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 arrayint 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 codeint 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
# Python3 program for Minimum swap required# to convert binary tree to binary search tree # Inorder Traversal of Binary Treedef 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 arraydef 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 Codeif __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
<script>// Javascript program for Minimum swap required// to convert binary tree to binary search tree // Inorder Traversal of Binary Treefunction 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 arrayfunction 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 codelet 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>
// Java program for Minimum swap required// to convert binary tree to binary search treeimport 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).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.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));    }}
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 write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
code_freak
mohit kumar 29
patel2127
adityapande88
saurabh1990aror
hardikkoriintern
Sorting
Tree
Sorting
Tree
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.