Dado un árbol binario y una cantidad de Nodes en el árbol, la tarea es encontrar la cantidad de pares que violan la propiedad BST . Binary Search Tree es una estructura de datos de árbol binario basada en Nodes que tiene las siguientes propiedades:
- El subárbol izquierdo de un Node contiene solo Nodes con claves menores que la clave del Node.
- El subárbol derecho de un Node contiene solo Nodes con claves mayores que la clave del Node.
- El subárbol izquierdo y derecho también debe ser un árbol de búsqueda binaria.
Ejemplos:
Input:
4
/ \
5 6
Output: 1
For the above binary tree, pair (5, 4)
violate the BST property. Thus, count
of pairs violating BST property is 1.
Input:
50
/ \
30 60
/ \ / \
20 25 10 40
Output: 7
For the above binary tree, pairs (20, 10),
(25, 10), (30, 25), (30, 10), (50, 10),
(50, 40), (60, 40) violate the BST property.
Thus, count of pairs violating BST property
is 7.
Acercarse:
- Almacene el recorrido en orden del árbol binario en una array.
- Ahora cuente todos los pares tales que a[i] > a[j] para i < j , que es el número de inversiones en la array.
- Imprime el conteo de pares que violan la propiedad BST .
A continuación se muestra la implementación del enfoque anterior:
Java
// Java program to count number of pairs
// in a binary tree violating the BST property
import java.io.*;
import java.util.*;
// Class that represents a node of the tree
class Node {
int data;
Node left, right;
// Constructor to create a new tree node
Node(int key)
{
data = key;
left = right = null;
}
}
class GFG {
// This method sorts the input array and returns the
// number of inversions in the array
static int mergeSort(int arr[], int array_size)
{
int temp[] = new int[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
// An auxiliary recursive method that sorts
// the input array and returns the number of
// inversions in the array
static int _mergeSort(int arr[], int temp[],
int left, int right)
{
int mid, inv_count = 0;
if (right > left) {
// Divide the array into two parts and
// call _mergeSortAndCountInv() for each
// of the parts
mid = (right + left) / 2;
// Inversion count will be sum of inversions
// in left-part, right-part and number of
// inversions in merging
inv_count = _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
// Merge the two parts
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
// This method merges two sorted arrays and returns
// inversion count in the arrays
static int merge(int arr[], int temp[], int left,
int mid, int right)
{
int i, j, k;
int inv_count = 0;
// i is index for left subarray
i = left;
// j is index for right subarray
j = mid;
// k is index for resultant merged subarray
k = left;
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
inv_count = inv_count + (mid - i);
}
}
// Copy the remaining elements of left subarray
// (if there are any) to temp
while (i <= mid - 1)
temp[k++] = arr[i++];
// Copy the remaining elements of right subarray
// if there are any) to temp
while (j <= right)
temp[k++] = arr[j++];
// Copy back the merged elements to original array
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
// Array to store
// inorder traversal of the binary tree
static int[] a;
static int in;
// Inorder traversal of the binary tree
static void Inorder(Node node)
{
if (node == null)
return;
Inorder(node.left);
a[in++] = node.data;
Inorder(node.right);
}
// Function to count the pairs
// in a binary tree violating BST property
static int pairsViolatingBST(Node root, int N)
{
if (root == null)
return 0;
in = 0;
a = new int[N];
Inorder(root);
// Total inversions in the array
int inversionCount = mergeSort(a, N);
return inversionCount;
}
// Driver code
public static void main(String args[])
{
int N = 7;
Node root = new Node(50);
root.left = new Node(30);
root.right = new Node(60);
root.left.left = new Node(20);
root.left.right = new Node(25);
root.right.left = new Node(10);
root.right.right = new Node(40);
System.out.println(pairsViolatingBST(root, N));
}
}
Python3
# Python3 program to count number of pairs # in a binary tree violating the BST property # Class that represents a node of the tree class newNode: def __init__(self, key): self.data = key self.left = None self.right = None # Array to store # inorder traversal of the binary tree a = [] id = 0 # This method sorts the input array # and returns the number of inversions # in the array def mergeSort(array_size): temp = [0] * array_size return _mergeSort(temp, 0, array_size - 1) # An auxiliary recursive method that sorts # the input array and returns the number of # inversions in the array def _mergeSort(temp, left, right): inv_count = 0 if (right > left): # Divide the array into two parts and # call _mergeSortAndCountInv() for each # of the parts mid = (right + left) // 2 # Inversion count will be sum of inversions # in left-part, right-part and number of # inversions in merging inv_count = _mergeSort(temp, left, mid) inv_count += _mergeSort(temp, mid + 1, right) # Merge the two parts inv_count += merge(temp, left, mid + 1, right) return inv_count # This method merges two sorted arrays # and returns inversion count in the arrays def merge(temp, left, mid, right): global a inv_count = 0 # i is index for left subarray i = left # j is index for right subarray j = mid # k is index for resultant merged subarray k = left while ((i <= mid - 1) and (j <= right)): if (a[i] <= a[j]): temp[k] = a[i] k += 1 i += 1 else: temp[k] = a[j] k += 1 j += 1 inv_count = inv_count + (mid - i) # Copy the remaining elements of left # subarray (if there are any) to temp while (i <= mid - 1): temp[k] = a[i] k += 1 i += 1 # Copy the remaining elements of right # subarray if there are any) to temp while (j <= right): temp[k] = a[j] k += 1 j += 1 # Copy back the merged elements # to original array for i in range(left, right + 1, 1): a[i] = temp[i] return inv_count # Inorder traversal of the binary tree def Inorder(node): global a global id if (node == None): return Inorder(node.left) a.append(node.data) id += 1 Inorder(node.right) # Function to count the pairs # in a binary tree violating # BST property def pairsViolatingBST(root, N): if (root == None): return 0 Inorder(root) # Total inversions in the array inversionCount = mergeSort(N) return inversionCount # Driver code if __name__ == '__main__': N = 7 root = newNode(50) root.left = newNode(30) root.right = newNode(60) root.left.left = newNode(20) root.left.right = newNode(25) root.right.left = newNode(10) root.right.right = newNode(40) print(pairsViolatingBST(root, N)) # This code is contributed by bgangwar59
C#
// C# program to count number of pairs
// in a binary tree violating the BST property
using System;
// Class that represents a node of the tree
public class Node {
public int data;
public Node left, right;
// Constructor to create a new tree node
public Node(int key)
{
data = key;
left = right = null;
}
}
class GFG {
// This method sorts the input array and returns the
// number of inversions in the array
static int mergeSort(int[] arr, int array_size)
{
int[] temp = new int[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
// An auxiliary recursive method that sorts
// the input array and returns the number of
// inversions in the array
static int _mergeSort(int[] arr, int[] temp,
int left, int right)
{
int mid, inv_count = 0;
if (right > left) {
// Divide the array into two parts and
// call _mergeSortAndCountInv() for each
// of the parts
mid = (right + left) / 2;
// Inversion count will be sum of inversions
// in left-part, right-part and number of
// inversions in merging
inv_count = _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
// Merge the two parts
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
// This method merges two sorted arrays and returns
// inversion count in the arrays
static int merge(int[] arr, int[] temp, int left,
int mid, int right)
{
int i, j, k;
int inv_count = 0;
// i is index for left subarray
i = left;
// j is index for right subarray
j = mid;
// k is index for resultant merged subarray
k = left;
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
inv_count = inv_count + (mid - i);
}
}
// Copy the remaining elements of left subarray
// (if there are any) to temp
while (i <= mid - 1)
temp[k++] = arr[i++];
// Copy the remaining elements of right subarray
// if there are any) to temp
while (j <= right)
temp[k++] = arr[j++];
// Copy back the merged elements to original array
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
// Array to store
// inorder traversal of the binary tree
static int[] a;
static int i;
// Inorder traversal of the binary tree
static void Inorder(Node node)
{
if (node == null)
return;
Inorder(node.left);
a[i++] = node.data;
Inorder(node.right);
}
// Function to count the pairs
// in a binary tree violating BST property
static int pairsViolatingBST(Node root, int N)
{
if (root == null)
return 0;
i = 0;
a = new int[N];
Inorder(root);
// Total inversions in the array
int inversionCount = mergeSort(a, N);
return inversionCount;
}
// Driver code
public static void Main(String[] args)
{
int N = 7;
Node root = new Node(50);
root.left = new Node(30);
root.right = new Node(60);
root.left.left = new Node(20);
root.left.right = new Node(25);
root.right.left = new Node(10);
root.right.right = new Node(40);
Console.WriteLine(pairsViolatingBST(root, N));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// JavaScript program to count number of pairs
// in a binary tree violating the BST property
// Class that represents a node of the tree
class Node
{
// Constructor to create a new tree node
constructor(key) {
this.data = key;
this.left = null;
this.right = null;
}
}
// This method sorts the input array and returns the
// number of inversions in the array
function mergeSort(arr, array_size) {
var temp = new Array(array_size).fill(0);
return _mergeSort(arr, temp, 0, array_size - 1);
}
// An auxiliary recursive method that sorts
// the input array and returns the number of
// inversions in the array
function _mergeSort(arr, temp, left, right) {
var mid,
inv_count = 0;
if (right > left)
{
// Divide the array into two parts and
// call _mergeSortAndCountInv() for each
// of the parts
mid = parseInt((right + left) / 2);
// Inversion count will be sum of inversions
// in left-part, right-part and number of
// inversions in merging
inv_count = _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
// Merge the two parts
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
// This method merges two sorted arrays and returns
// inversion count in the arrays
function merge(arr, temp, left, mid, right) {
var i, j, k;
var inv_count = 0;
// i is index for left subarray
i = left;
// j is index for right subarray
j = mid;
// k is index for resultant merged subarray
k = left;
while (i <= mid - 1 && j <= right) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
} else {
temp[k++] = arr[j++];
inv_count = inv_count + (mid - i);
}
}
// Copy the remaining elements of left subarray
// (if there are any) to temp
while (i <= mid - 1) temp[k++] = arr[i++];
// Copy the remaining elements of right subarray
// if there are any) to temp
while (j <= right) temp[k++] = arr[j++];
// Copy back the merged elements to original array
for (i = left; i <= right; i++) arr[i] = temp[i];
return inv_count;
}
// Array to store
// inorder traversal of the binary tree
var a = [];
var i;
// Inorder traversal of the binary tree
function Inorder(node) {
if (node == null) return;
Inorder(node.left);
a[i++] = node.data;
Inorder(node.right);
}
// Function to count the pairs
// in a binary tree violating BST property
function pairsViolatingBST(root, N) {
if (root == null) return 0;
i = 0;
a = new Array(N).fill(0);
Inorder(root);
// Total inversions in the array
var inversionCount = mergeSort(a, N);
return inversionCount;
}
// Driver code
var N = 7;
var root = new Node(50);
root.left = new Node(30);
root.right = new Node(60);
root.left.left = new Node(20);
root.left.right = new Node(25);
root.right.left = new Node(10);
root.right.right = new Node(40);
document.write(pairsViolatingBST(root, N) + "<br>");
// This code is contributed by rdtank.
</script>
Producción:
7
Publicación traducida automáticamente
Artículo escrito por rachana soma y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA