Minimum swap required to convert binary tree to binary search tree

Last Updated : 25 Nov, 2024

Given an array arr[] which represents a 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 swaps required to convert it into a Binary Search Tree.

Examples:

Input: arr[] = [5, 6, 7, 8, 9, 10, 11]
Output: 3
Explanation:
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 to obtain the below binary search tree:

Input: arr[] = [1, 2, 3]
Output: 1
Explanation:
Binary tree of the given array:

After swapping node 1 with node 2, obtain the below binary search tree:

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.

C++

// C++ program for Minimum swap required // to convert binary tree to binary search tree  #include<bits/stdc++.h> using namespace std;  // Function to perform inorder traversal of the binary tree // and store it in vector v void inorder(vector<int>& arr, vector<int>& inorderArr, int index) {        int n = arr.size();          // If index is out of bounds, return     if (index >= n)         return;      // Recursively visit left subtree     inorder(arr, inorderArr, 2 * index + 1);          // Store current node value in vector     inorderArr.push_back(arr[index]);          // Recursively visit right subtree     inorder(arr, inorderArr, 2 * index + 2); }  // Function to calculate minimum swaps  // to sort inorder traversal int minSwaps(vector<int>& arr) {     int n = arr.size();     vector<int> inorderArr;          // Get the inorder traversal of the binary tree     inorder(arr, inorderArr, 0);          // Create an array of pairs to store value   	// and original index     vector<pair<int, int>> t(inorderArr.size());     int ans = 0;          // Store the value and its index     for (int i = 0; i < inorderArr.size(); i++)         t[i] = {inorderArr[i], i};          // Sort the pair array based on values    	// to get BST order     sort(t.begin(), t.end());          // Find minimum swaps by detecting cycles     for (int i = 0; i < t.size(); i++) {                // If the element is already in the        	// correct position, continue         if (i == t[i].second)             continue;                  // Otherwise, perform swaps until the element       	// is in the right place         else {                        // Swap elements to correct positions             swap(t[i].first, t[t[i].second].first);             swap(t[i].second, t[t[i].second].second);         }                  // Check if the element is still not       	// in the correct position         if (i != t[i].second)             --i;                    // Increment swap count         ans++;     }        return ans; }  int main() {        vector<int> arr = { 5, 6, 7, 8, 9, 10, 11 };     cout << minSwaps(arr) << endl; }

Java

// Java program for Minimum swap required // to convert binary tree to binary search tree import java.util.Arrays;  class GfG {          // Function to perform inorder traversal of the binary tree     // and store it in an array     static void inorder(int[] arr, int[] inorderArr,                          int index, int[] counter) {         int n = arr.length;                  // Base case: if index is out of bounds, return         if (index >= n)             return;                  // Recursively visit left subtree         inorder(arr, inorderArr, 2 * index + 1, counter);                  // Store current node value in the inorder array         inorderArr[counter[0]] = arr[index];         counter[0]++;                  // Recursively visit right subtree         inorder(arr, inorderArr, 2 * index + 2, counter);     }      // Function to calculate minimum swaps      // to sort inorder traversal     static int minSwaps(int[] arr) {         int n = arr.length;         int[] inorderArr = new int[n];         int[] counter = new int[1];                  // Get the inorder traversal of the binary tree         inorder(arr, inorderArr, 0, counter);                  // Create an array of pairs to store the value          // and its original index         int[][] t = new int[n][2];         int ans = 0;                  // Store the value and its original index         for (int i = 0; i < n; i++) {             t[i][0] = inorderArr[i];             t[i][1] = i;         }                  // Sort the array based on values to get BST order         Arrays.sort(t, (a, b) -> Integer.compare(a[0], b[0]));                  // Find minimum swaps by detecting cycles         boolean[] visited = new boolean[n];                  // Iterate through the array to find cycles         for (int i = 0; i < n; i++) {                        // If the element is already visited or in           	// the correct place, continue             if (visited[i] || t[i][1] == i)                 continue;                          // Start a cycle and find the number of           	// nodes in the cycle             int cycleSize = 0;             int j = i;                          while (!visited[j]) {                 visited[j] = true;                 j = t[j][1];                 cycleSize++;             }                          // If there is a cycle, we need (cycleSize - 1)           	// swaps to sort the cycle             if (cycleSize > 1) {                 ans += (cycleSize - 1);             }         }                  // Return the total number of swaps         return ans;     }      public static void main(String[] args) {         int[] arr = {5, 6, 7, 8, 9, 10, 11};          System.out.println(minSwaps(arr));     } }

Python

# Python program for Minimum swap required # to convert binary tree to binary search tree  # Function to perform inorder traversal of the binary tree # and store it in an array def inorder(arr, inorderArr, index):        # If index is out of bounds, return     n = len(arr)     if index >= n:         return      # Recursively visit left subtree     inorder(arr, inorderArr, 2 * index + 1)      # Store current node value in inorderArr     inorderArr.append(arr[index])      # Recursively visit right subtree     inorder(arr, inorderArr, 2 * index + 2)  # Function to calculate minimum swaps  # to sort inorder traversal def minSwaps(arr):     inorderArr = []      # Get the inorder traversal of the binary tree     inorder(arr, inorderArr, 0)      # Create a list of pairs to store value and original index     t = [(inorderArr[i], i) for i in range(len(inorderArr))]     ans = 0      # Sort the list of pairs based on values     # to get BST order     t.sort()      # Initialize visited array     visited = [False] * len(t)      # Find minimum swaps by detecting cycles     for i in range(len(t)):          # If already visited or already in the         # correct place, skip         if visited[i] or t[i][1] == i:             continue          # Start a cycle and find the number of          # nodes in the cycle         cycleSize = 0         j = i          # Process all elements in the cycle         while not visited[j]:             visited[j] = True             j = t[j][1]             cycleSize += 1          # If there is a cycle of size `cycle_size`, we          # need `cycle_size - 1` swaps         if cycleSize > 1:             ans += (cycleSize - 1)      # Return total number of swaps     return ans  if __name__ == "__main__":     arr = [5, 6, 7, 8, 9, 10, 11]     print(minSwaps(arr))

// C# program for Minimum swap required // to convert binary tree to binary search tree using System; using System.Linq;  class GfG {        // Function to perform inorder traversal of the binary tree     // and store it in an array     static void Inorder(int[] arr, int[] inorderArr, int index, ref int counter) {         int n = arr.Length;          // Base case: if index is out of bounds, return         if (index >= n)             return;          // Recursively visit left subtree         Inorder(arr, inorderArr, 2 * index + 1, ref counter);          // Store current node value in inorderArr         inorderArr[counter] = arr[index];         counter++;          // Recursively visit right subtree         Inorder(arr, inorderArr, 2 * index + 2, ref counter);     }      // Function to calculate minimum   	// swaps to sort inorder traversal     static int MinSwaps(int[] arr) {         int n = arr.Length;         int[] inorderArr = new int[n];         int counter = 0;          // Get the inorder traversal of the binary tree         Inorder(arr, inorderArr, 0, ref counter);          // Create an array of pairs to store value        	// and original index         var t = new (int, int)[n];         for (int i = 0; i < n; i++) {             t[i] = (inorderArr[i], i);         }          // Sort the array based on values to get BST order         Array.Sort(t, (a, b) => a.Item1.CompareTo(b.Item1));          // Initialize visited array         bool[] visited = new bool[n];         int ans = 0;          // Find minimum swaps by detecting cycles         for (int i = 0; i < n; i++) {                        // If already visited or already in            	// the correct place, skip             if (visited[i] || t[i].Item2 == i)                 continue;              // Start a cycle and find the number            	// of nodes in the cycle             int cycleSize = 0;             int j = i;              // Process all elements in the cycle             while (!visited[j]) {                 visited[j] = true;                 j = t[j].Item2;                 cycleSize++;             }              // If there is a cycle of size `cycle_size`, we          	// need `cycle_size - 1` swaps             if (cycleSize > 1)             {                 ans += (cycleSize - 1);             }         }          // Return total number of swaps         return ans;     }      static void Main(string[] args) {                 int[] arr = { 5, 6, 7, 8, 9, 10, 11 };         Console.WriteLine(MinSwaps(arr));     } }

JavaScript

// Javascript program for Minimum swap required // to convert binary tree to binary search tree  // Inorder traversal to get values in sorted order function inorder(arr, inorderArr, index) {      // If index is out of bounds, return     if (index >= arr.length)         return;      // Recursively visit left subtree     inorder(arr, inorderArr, 2 * index + 1);      // Store current node value in array     inorderArr.push(arr[index]);      // Recursively visit right subtree     inorder(arr, inorderArr, 2 * index + 2); }  // Function to calculate minimum swaps to sort inorder // traversal function minSwaps(arr) {      let inorderArr = [];      // Get the inorder traversal of the binary tree     inorder(arr, inorderArr, 0);      // Create an array of pairs to store value and original     // index     let t = inorderArr.map((val, i) => [val, i]);     let ans = 0;      // Sort the pair array based on values to get BST order     t.sort((a, b) => a[0] - b[0]);      // Find minimum swaps by detecting cycles     let visited = Array(arr.length)                       .fill(false);      for (let i = 0; i < t.length; i++) {              // If the element is already in the correct         // position, continue         if (visited[i] || t[i][1] === i)             continue;          // Otherwise, perform swaps until the element is in         // the right place         let cycleSize = 0;         let j = i;          while (!visited[j]) {             visited[j] = true;             j = t[j][1];             cycleSize++;         }          // If there is a cycle, we need (cycleSize - 1)         // swaps to sort the cycle         if (cycleSize > 1) {             ans += (cycleSize - 1);         }     }      // Return total number of swaps     return ans; }  let arr = [ 5, 6, 7, 8, 9, 10, 11 ]; console.log(minSwaps(arr));

Output

Time Complexity: O(n*logn) where n is the number of elements in array.
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?

Convert Binary Tree to Doubly Linked List using inorder traversal

Anuj Chauhan

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Minimum swap required to convert binary tree to binary search tree

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