Count of pairs violating BST property

Given a Binary tree and number of nodes in the tree, the task is to find the number of pairs violating the BST property. Binary Search Tree is a node-based binary tree data structure which has the following properties:

  • The left subtree of a node contains only nodes with keys lesser than the node’s key.
  • The right subtree of a node contains only nodes with keys greater than the node’s key.
  • The left and right subtree each must also be a binary search tree.

Examples:

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.

Approach:

  • Store the inorder traversal of the binary tree in an array.
  • Now count all the pairs such that a[i] > a[j] for i < j which is number of inversions in the array.
  • Print the count of pairs violating BST property.

Below is the implementation of the above approach:

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// 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));
    }
}

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Output:

7


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