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Es requereix un intercanvi mínim per convertir l'arbre binari a l'arbre de cerca binari

Donada una matriu arr[] que representa a Arbre binari complet és a dir, si índex i és el pare índex 2*i + 1 és el fill deixat i índex 2*i + 2 és el nen adequat. La tasca és trobar el mínim nombre de intercanvis necessari per convertir-lo en a Arbre de cerca binari.

Exemples:  

Entrada: arr[] = [5 6 7 8 9 10 11]
Sortida: 3
Explicació:
Arbre binari de la matriu donada:



Canvi mínim necessari per convertir l'arbre binari a l'arbre de cerca binari 1' title=

Canvi 1: Canvia el node 8 amb el node 5.
Canvi 2: Canvia el node 9 amb el node 10.
Intercanvi 3: intercanvieu el node 10 amb el node 7.

Per tant, calen un mínim de 3 intercanvis per obtenir l'arbre de cerca binari següent:

comunicació analògica
Canvi mínim necessari per convertir l'arbre binari a l'arbre de cerca binari 3' loading='lazy' title=


Entrada: arr[] = [1 2 3]
Sortida: 1
Explicació:
Arbre binari de la matriu donada:

Canvi mínim necessari per convertir l'arbre binari a l'arbre de cerca binari 2' loading='lazy' title=

Després d'intercanviar el node 1 amb el node 2, obteniu l'arbre de cerca binari següent:

Canvi mínim necessari per convertir l'arbre binari a l'arbre de cerca binari 4' loading='lazy' title=

Enfocament:

marca d'aigua en word

La idea és utilitzar el fet que travessa en ordre de Arbre de cerca binari està dins augmentant ordre del seu valor. 
Així que troba el travessa en ordre de l'Arbre Binari i emmagatzemar-lo a la matriu i intentar-ho ordenar la matriu. El nombre mínim d'intercanvis necessaris per ordenar la matriu serà la resposta.

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)); 

Sortida
3 

Complexitat temporal: O(n*logn) on n és el nombre d'elements de la matriu.
Espai auxiliar: O(n) perquè utilitza espai addicional per a la matriu 

Exercici: Podem estendre això a l'arbre binari normal, és a dir, un arbre binari representat amb punters esquerre i dret i no necessàriament complet?

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