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# Two nodes of a BST are swapped, correct the BST

• Difficulty Level : Hard
• Last Updated : 17 Jun, 2021

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++

 `// Two nodes in the BST's swapped, correct the BST.``#include ``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;``};` `// 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 pointers 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 Inorder traversal */``void` `printInorder(``struct` `node* node)``{``    ``if` `(node == NULL)``        ``return``;``    ``printInorder(node->left);``    ``cout <<``"  "``<< 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);` `    ``cout <<``"Inorder Traversal of the original tree \n"``;``    ``printInorder(root);` `    ``correctBST(root);` `    ``cout <<``"\nInorder Traversal of the fixed tree \n"``;``    ``printInorder(root);` `    ``return` `0;``}` `// This code is contributed by shivanisinghss2110`

## C

 `// Two nodes in the BST's swapped, correct the BST.``#include ``#include ` `/* 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 pointers 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 Inorder 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;``}`

## Java

 `// 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 pointers 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``     ``Inorder 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`

## Python3

 `# 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 required 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`

## C#

 `// 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 pointers 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`

## Javascript

 ``

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.