Two nodes of a BST are swapped, correct the BST

Two of the nodes of a Binary Search Tree (BST) are swapped. Fix (or correct) the BST. 

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

The inorder traversal of a BST produces a sorted array. So a simple method is to store inorder traversal of the input tree in an auxiliary array. Sort the auxiliary array. Finally, insert the auxiliary array elements back to the BST, keeping the structure of the BST same. The time complexity of this method is O(nLogn) and the auxiliary space needed is O(n).
We can solve this in O(n) time and with a single traversal of the given BST. Since inorder traversal of BST is always a sorted array, the problem can be reduced to a problem where two elements of a sorted array are swapped. There are two cases that we need to handle:
1. The swapped nodes are not adjacent in the inorder traversal of the BST. 

 For example, Nodes 5 and 25 are swapped in {3 5 7 8 10 15 20 25}. 
 The inorder traversal of the given tree is 3 25 7 8 10 15 20 5 

If we observe carefully, during inorder traversal, we find node 7 is smaller than the previous visited node 25. Here save the context of node 25 (previous node). Again, we find that node 5 is smaller than the previous node 20. This time, we save the context of node 5 (the current node ). Finally, swap the two node’s values.
2. The swapped nodes are adjacent in the inorder traversal of BST.

  For example, Nodes 7 and 8 are swapped in {3 5 7 8 10 15 20 25}. 
  The inorder traversal of the given tree is 3 5 8 7 10 15 20 25 

Unlike case #1, here only one point exists where a node value is smaller than the previous node value. e.g. node 7 is smaller than node 8. 
How to Solve? We will maintain three-pointers, first, middle, and last. When we find the first point where the current node value is smaller than the previous node value, we update the first with the previous node & the middle with the current node. When we find the second point where the current node value is smaller than the previous node value, we update the last with the current node. In the case of #2, we will never find the second point. So, the last pointer will not be updated. After processing, if the last node value is null, then two swapped nodes of BST are adjacent
Following is the implementation of the given code.
 

C++

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// Two nodes in the BST's swapped, correct the BST.
#include <stdio.h>
#include <stdlib.h>
  
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node
{
    int data;
    struct node *left, *right;
};
  
// A utility function to swap two integers
void swap( int* a, int* b )
{
    int t = *a;
    *a = *b;
    *b = t;
}
  
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(int data)
{
    struct node* node = (struct node *)malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
    return(node);
}
  
// This function does inorder traversal to find out the two swapped nodes.
// It sets three pointers, first, middle and last.  If the swapped nodes are
// adjacent to each other, then first and middle contain the resultant nodes
// Else, first and last contain the resultant nodes
void correctBSTUtil( struct node* root, struct node** first,
                     struct node** middle, struct node** last,
                     struct node** prev )
{
    if( root )
    {
        // Recur for the left subtree
        correctBSTUtil( root->left, first, middle, last, prev );
  
        // If this node is smaller than the previous node, it's violating
        // the BST rule.
        if (*prev && root->data < (*prev)->data)
        {
            // If this is first violation, mark these two nodes as
            // 'first' and 'middle'
            if ( !*first )
            {
                *first = *prev;
                *middle = root;
            }
  
            // If this is second violation, mark this node as last
            else
                *last = root;
        }
  
        // Mark this node as previous
        *prev = root;
  
        // Recur for the right subtree
        correctBSTUtil( root->right, first, middle, last, prev );
    }
}
  
// A function to fix a given BST where two nodes are swapped.  This
// function uses correctBSTUtil() to find out two nodes and swaps the
// nodes to fix the BST
void correctBST( struct node* root )
{
    // Initialize pointers needed for correctBSTUtil()
    struct node *first, *middle, *last, *prev;
    first = middle = last = prev = NULL;
  
    // Set the poiters to find out two nodes
    correctBSTUtil( root, &first, &middle, &last, &prev );
  
    // Fix (or correct) the tree
    if( first && last )
        swap( &(first->data), &(last->data) );
    else if( first && middle ) // Adjacent nodes swapped
        swap( &(first->data), &(middle->data) );
  
    // else nodes have not been swapped, passed tree is really BST.
}
  
/* 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);
}
  
/* Driver program to test above functions*/
int main()
{
    /*   6
        /  \
       10    2
      / \   / \
     1   3 7  12
     10 and 2 are swapped
    */
  
    struct node *root = newNode(6);
    root->left        = newNode(10);
    root->right       = newNode(2);
    root->left->left  = newNode(1);
    root->left->right = newNode(3);
    root->right->right = newNode(12);
    root->right->left = newNode(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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Java

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// Java program to correct the BST 
// if two nodes are swapped
import java.util.*;
import java.lang.*;
import java.io.*;
  
class Node {
  
    int data;
    Node left, right;
  
    Node(int d) {
        data = d;
        left = right = null;
    }
}
  
class BinaryTree
{
    Node first, middle, last, prev;
      
    // This function does inorder traversal
    // to find out the two swapped nodes.
    // It sets three pointers, first, middle
    // and last. If the swapped nodes are
    // adjacent to each other, then first 
    // and middle contain the resultant nodes
    // Else, first and last contain the 
    // resultant nodes
    void correctBSTUtil( Node root)
    {
        if( root != null )
        {
            // Recur for the left subtree
            correctBSTUtil( root.left);
  
            // If this node is smaller than
            // the previous node, it's 
            // violating the BST rule.
            if (prev != null && root.data <
                                prev.data)
            {
                // If this is first violation,
                // mark these two nodes as
                // 'first' and 'middle'
                if (first == null)
                {
                    first = prev;
                    middle = root;
                }
  
                // If this is second violation,
                // mark this node as last
                else
                    last = root;
            }
  
            // Mark this node as previous
            prev = root;
  
            // Recur for the right subtree
            correctBSTUtil( root.right);
        }
    }
  
    // A function to fix a given BST where 
    // two nodes are swapped. This function 
    // uses correctBSTUtil() to find out 
    // two nodes and swaps the nodes to 
    // fix the BST
    void correctBST( Node root )
    {
        // Initialize pointers needed 
        // for correctBSTUtil()
        first = middle = last = prev = null;
  
        // Set the poiters to find out 
        // two nodes
        correctBSTUtil( root );
  
        // Fix (or correct) the tree
        if( first != null && last != null )
        {
            int temp = first.data;
            first.data = last.data;
            last.data = temp; 
        }
        // Adjacent nodes swapped
        else if( first != null && middle !=
                                    null
        {
            int temp = first.data;
            first.data = middle.data;
            middle.data = temp;
        }
  
        // else nodes have not been swapped,
        // passed tree is really BST.
    }
  
    /* A utility function to print 
     Inoder traversal */
    void printInorder(Node node)
    {
        if (node == null)
            return;
        printInorder(node.left);
        System.out.print(" " + node.data);
        printInorder(node.right);
    }
  
  
    // Driver program to test above functions
    public static void main (String[] args) 
    {
        /*   6
            / \
           10  2
          / \ / \
         1  3 7 12
          
        10 and 2 are swapped
        */
  
        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);
  
        System.out.println("Inorder Traversal"+
                        " of the original tree");
        BinaryTree tree = new BinaryTree();
        tree.printInorder(root);
  
        tree.correctBST(root);
  
        System.out.println("\nInorder Traversal"+
                          " of the fixed tree");
        tree.printInorder(root);
    }
}
// This code is contributed by Chhavi

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Python3

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# Python3 program to correct the BST  
# if two nodes are swapped 
class Node: 
      
    # Constructor to create a new node 
    def __init__(self, data):
          
        self.key = data 
        self.left = None
        self.right = None
  
# Utility function to track the nodes
# that we have to swap
def correctBstUtil(root, first, middle,
                   last, prev):
                         
    if(root):
          
        # Recur for the left sub tree
        correctBstUtil(root.left, first,
                       middle, last, prev)
                         
        # If this is the first violation, mark these 
        # two nodes as 'first and 'middle'
        if(prev[0] and root.key < prev[0].key):
            if(not first[0]):
                first[0] = prev[0]
                middle[0] = root
            else:
                  
                # If this is the second violation,
                # mark this node as last
                last[0] = root
                  
        prev[0] = root
          
        # Recur for the right subtree
        correctBstUtil(root.right, first, 
                       middle, last, prev)
      
# A function to fix a given BST where 
# two nodes are swapped. This function
# uses correctBSTUtil() to find out two
# nodes and swaps the nodes to fix the BST 
def correctBst(root):
      
    # Followed four lines just for forming
    # an array with only index 0 filled 
    # with None and we will update it accordingly.
    # we made it null so that we can fill 
    # node data in them.
    first = [None]
    middle = [None]
    last = [None]
    prev = [None]
      
    # Setting arrays (having zero index only) 
    # for capturing the requird node
    correctBstUtil(root, first, middle,
                   last, prev)
  
    # Fixing the two nodes
    if(first[0] and last[0]):
          
        # Swapping for first and last key values
        first[0].key, last[0].key = (last[0].key, 
                                    first[0].key)
  
    elif(first[0] and middle[0]):
      
        # Swapping for first and middle key values
        first[0].key, middle[0].key = (middle[0].key,
                                        first[0].key)
      
    # else tree will be fine
  
# Function to print inorder
# traversal of tree
def PrintInorder(root):
      
    if(root):
        PrintInorder(root.left)
        print(root.key, end = " ")
        PrintInorder(root.right)
          
    else:
        return
  
# Driver code
  
#      6
#     /   \
#   10    2
#  / \   / \
# 1   3 7   12
  
# Following 7 lines are for tree formation
root = Node(6
root.left = Node(10
root.right = Node(2
root.left.left = Node(1
root.left.right = Node(3
root.right.left = Node(7
root.right.right = Node(12
  
# Printing inorder traversal of normal tree
print("inorder traversal of noraml tree")
PrintInorder(root)
print("")
  
# Function call to do the task
correctBst(root)
  
# Printing inorder for corrected Bst tree
print("")
print("inorder for corrected BST")
  
PrintInorder(root)
  
# This code is contributed by rajutkarshai

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C#

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// C# program to correct the BST 
// if two nodes are swapped
using System;
class Node{
  
public int data;
public Node left, right;
public Node(int d) 
{
  data = d;
  left = right = null;
}
}
  
class BinaryTree
{
  Node first, middle, 
       last, prev;
  
// This function does inorder traversal
// to find out the two swapped nodes.
// It sets three pointers, first, middle
// and last. If the swapped nodes are
// adjacent to each other, then first 
// and middle contain the resultant nodes
// Else, first and last contain the 
// resultant nodes
void correctBSTUtil( Node root)
{
  if( root != null )
  {
    // Recur for the 
    // left subtree
    correctBSTUtil(root.left);
  
    // If this node is smaller than
    // the previous node, it's 
    // violating the BST rule.
    if (prev != null && root.data <
        prev.data)
    {
      // If this is first violation,
      // mark these two nodes as
      // 'first' and 'middle'
      if (first == null)
      {
        first = prev;
        middle = root;
      }
  
      // If this is second violation,
      // mark this node as last
      else
        last = root;
    }
  
    // Mark this node 
    // as previous
    prev = root;
  
    // Recur for the
    // right subtree
    correctBSTUtil(root.right);
  }
}
  
// A function to fix a given BST where 
// two nodes are swapped. This function 
// uses correctBSTUtil() to find out 
// two nodes and swaps the nodes to 
// fix the BST
void correctBST( Node root )
{
  // Initialize pointers needed 
  // for correctBSTUtil()
  first = middle = last = 
          prev = null;
  
  // Set the poiters to 
  // find out two nodes
  correctBSTUtil(root);
  
  // Fix (or correct) 
  // the tree
  if(first != null && 
     last != null)
  {
    int temp = first.data;
    first.data = last.data;
    last.data = temp; 
  }
    
  // Adjacent nodes swapped
  else if(first != null && 
          middle != null
  {
    int temp = first.data;
    first.data = middle.data;
    middle.data = temp;
  }
  
  // else nodes have not been 
  // swapped, passed tree is 
  // really BST.
}
  
// A utility function to print 
// Inoder traversal 
void printInorder(Node node)
{
  if (node == null)
    return;
  printInorder(node.left);
  Console.Write(" " + node.data);
  printInorder(node.right);
}
  
// Driver code
public static void Main(String[] args) 
{
  /*         6
            / \
           10  2
          / \ / \
         1  3 7 12
  
        10 and 2 are swapped
        */
  
  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);
  
  Console.WriteLine("Inorder Traversal" +
                    " of the original tree");
  BinaryTree tree = new BinaryTree();
  tree.printInorder(root);
  tree.correctBST(root);
  Console.WriteLine("\nInorder Traversal" +
                    " of the fixed tree");
  tree.printInorder(root);
}
}
  
// This code is contributed by gauravrajput1

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Output: 

Inorder Traversal of the original tree
1 10 3 6 7 2 12
Inorder Traversal of the fixed tree
1 2 3 6 7 10 12






Time Complexity: O(n)


See this for different test cases of the above code.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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