Two nodes of a BST are swapped, correct the BST | Set-2

Given a Binary Search Tree with two of the nodes of the Binary Search Tree (BST) swapped. The task is to fix (or correct) the BST.

Note: The BST will not have duplicates.

Examples:



Input Tree:
         10
        /  \
       5    8
      / \
     2   20

In the above tree, nodes 20 and 8 must be swapped to fix the tree.  
Following is the output tree
         10
        /  \
       5    20
      / \
     2   8

Approach :

  • Traverse the BST in In-order fashion and store the nodes in a vector.
  • Then this vector is sorted after creating a copy of it.
  • Use Insertion sort as it works the best and fastest when the array is almost sorted. As in this problem, only two elements will be displaced so Insertion sort here will work in linear time.
  • After sorting, compare the sorted vector and the copy of the vector created earlier, by this, find out the error-some nodes and fix them by finding them in the tree and exchanging them.

Below is the implementation of above approach:

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
   
// A binary tree node has data, pointer 
// to left child and a pointer to right child 
struct node {
    int data;
    struct node *left, *right;
    node(int x)
    {
        data = x;
        left = right = NULL;
    }
};
   
// Utility function for insertion sort
void insertionSort(vector<int>& v, int n)
{
    int i, key, j;
    for (i = 1; i < n; i++) {
        key = v[i];
        j = i - 1;
        while (j >= 0 && v[j] > key) {
            v[j + 1] = v[j];
            j = j - 1;
        }
        v[j + 1] = key;
    }
}
   
// Utility function to create a vector 
// with inorder traversal of a binary tree
void inorder(node* root, vector<int>& v)
{
    // Base cases
    if (!root)
        return;
   
    // Recurive call for left subtree
    inorder(root->left, v);
   
    // Insert node into vector
    v.push_back(root->data);
   
    // Recursive call for right subtree
    inorder(root->right, v);
}
   
// Function to exchange data 
// for incorrect nodes
void find(node* root, int res, int res2)
{
    // Base cases
    if (!root) {
        return;
    }
   
    // Recurive call to find 
    // the node in left subtree
    find(root->left, res, res2);
   
    // Check if current node 
    // is incorrect and exchange
    if (root->data == res) {
        root->data = res2;
    }
    else if (root->data == res2) {
        root->data = res;
    }
   
    // Recurive call to find 
    // the node in right subtree
    find(root->right, res, res2);
}
   
// Primary function to fix the two nodes
struct node* correctBST(struct node* root)
{
    // Vector to store the 
    // inorder traversal of tree
    vector<int> v;
   
    // Function call to insert
    // nodes into vector
    inorder(root, v);
   
    // create a copy of the vector
    vector<int> v1 = v;
   
    // Sort the original vector thereby 
    // making it a valid BST's inorder
    insertionSort(v, v.size());
   
    // Traverse through both vectors and
    // compare misplaced values in original BST
    for (int i = 0; i < v.size(); i++) {
  
        // Find the mismatched values
        // and interchange them
        if (v[i] != v1[i]) {
  
            // Find and exchange the 
            // data of the two nodes
            find(root, v1[i], v[i]);
   
            // As it given only two values are 
            // wrong we don't need to check further
            break;
        }
    }
   
    // Return the root of corrected BST
    return root;
}
  
// A utility function to
// print Inoder traversal 
void printInorder(struct node* node)
{
    if (node == NULL)
        return;
    printInorder(node->left);
    printf("%d ", node->data);
    printInorder(node->right);
}
   
int main()
{ 
    struct node* root = new node(6);
    root->left = new node(10);
    root->right = new node(2);
    root->left->left = new node(1);
    root->left->right = new node(3);
    root->right->right = new node(12);
    root->right->left = new node(7);
   
    printf("Inorder Traversal of the
                           original tree \n");
  
    printInorder(root);
   
    correctBST(root);
   
    printf("\nInorder Traversal of the 
                              fixed tree \n");
  
    printInorder(root);
   
    return 0;
}

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Time Complexity: O(N)

Auxiliary Space: O(N)

where N is the number of nodes in the Binary Tree.



          
          
          
          

            

    

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