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Intercambio mínimo requerido para convertir un árbol binario en un árbol de búsqueda binario

Dada una matriz llegar[] que representa un Árbol binario completo es decir, si índice i es el padre índice 2*yo + 1 es el niño abandonado y índice 2*i + 2 es el niño correcto. La tarea es encontrar el mínimo numero de permutas necesaria para convertirlo en un Árbol de búsqueda binaria.

Ejemplos:  

Aporte: arreglo[] = [5 6 7 8 9 10 11]
Producción: 3
Explicación:
Árbol binario de la matriz dada:



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Intercambio 1: intercambie el nodo 8 con el nodo 5.
Intercambio 2: intercambie el nodo 9 con el nodo 10.
Intercambio 3: intercambie el nodo 10 con el nodo 7.

Por lo tanto, se requieren un mínimo de 3 intercambios para obtener el siguiente árbol de búsqueda binaria:

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Aporte: arreglo[] = [1 2 3]
Producción: 1
Explicación:
Árbol binario de la matriz dada:

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Después de intercambiar el nodo 1 con el nodo 2, obtenga el siguiente árbol de búsqueda binario:

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Acercarse:

La idea es aprovechar el hecho de que recorrido en orden de Árbol de búsqueda binaria esta en creciente orden de su valor. 
Así que encuentra el recorrido en orden del árbol binario y guárdelo en la matriz y tratar de clasificar la matriz. El número mínimo de intercambios necesarios para ordenar la matriz será la respuesta.

C++
// C++ program for Minimum swap required // to convert binary tree to binary search tree #include   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#
// 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)); 

Producción
3 

Complejidad del tiempo: O(n*logn) donde norte es el número de elementos de la matriz.
Espacio Auxiliar: O(n) porque está usando espacio adicional para la matriz 

Ejercicio: ¿Podemos extender esto al árbol binario normal, es decir, un árbol binario representado mediante punteros izquierdo y derecho y no necesariamente completo?

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