# 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 the above approach:

 `// C++ implementation of the above approach` `#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;` `    ``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"``);` `    ``printf``(``"original tree \n"``);`   `    ``printInorder(root);` ` `  `    ``correctBST(root);` ` `  `    ``printf``(``"\nInorder Traversal of the"``); ` `    ``printf``(``"fixed tree \n"``);`   `    ``printInorder(root);` ` `  `    ``return` `0;` `}`

Output
```Inorder Traversal of theoriginal tree
1 10 3 6 7 2 12
Inorder Traversal of thefixed tree
1 2 3 6 7 10 12

```

Time Complexity: O(N)
Auxiliary Space: O(N), where N is the number of nodes in the Binary Tree.

Method 2:

To understand this, you need to first understand Morris Traversal or Morris Threading Traversal. It makes use of leaf nodes’ right/left pointer to achieve O(1) space Traversal on a Binary Tree.

So, in this approach, we can solve this in O(n) time and O(1) space i.e constant space, with a single traversal of the given BST. Since the 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.

Below is the implementation of the above approach:

 `// C++ implementation of the` `// above approach`   `#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); ` `}`   `// Function for inorder traversal using ` `// Morris Traversal` `void` `MorrisTraversal(``struct` `node* root,` `                     ``struct` `node* &first,` `                     ``struct` `node* &last,` `                     ``struct` `node* &prev ) ` `{ ` `       ``// Current node` `      ``struct`  `node* curr = root;` `    `  `        ``while``(curr != NULL)` `        ``{` `            ``if``(curr->left==NULL)` `            ``{` `              `  `                ``// If this node is smaller than ` `                ``// the previous node, it's  ` `                ``// violating the BST rule. ` `                ``if``(first == NULL && prev != NULL && ` `                   ``prev->data > curr->data)` `                ``{` `                    ``// If this is first violation, ` `                    ``// mark these two nodes as ` `                    ``// 'first' and 'last'` `                    ``first = prev;` `                    ``last = curr;` `                ``}` `                `  `                ``if``(first != NULL && ` `                   ``prev->data > curr->data)` `                ``{` `                  ``// If this is second violation,          ` `                  ``// mark this node as last` `                    ``last = curr;` `                ``}` `                ``prev = curr;` `                `  `                ``curr = curr->right;` `            ``}` `            ``else` `            ``{` `                  ``/* Find the inorder predecessor of current */` `                 ``struct`   `node*  pre = curr->left; ` `                ``while``(pre->right!=NULL && pre->right!=curr)` `                ``{` `                    ``pre = pre->right;` `                ``}` `                `  `                ``/* Make current as right child of ` `                ``its inorder predecessor */` `                ``if``(pre->right==NULL)` `                ``{` `                    ``pre->right = curr;` `                    ``curr = curr->left;` `                ``}` `                ``else` `                ``{` `                    ``// If this node is smaller than ` `                    ``// the previous node, it's  ` `                    ``// violating the BST rule. ` `                    ``if``(first == NULL && prev!=NULL &&` `                       ``prev->data > curr->data)` `                    ``{` `                          ``// If this is first violation, ` `                        ``// mark these two nodes as ` `                        ``// 'first' and 'last' ` `                        ``first = prev;` `                        ``last = curr;` `                    ``}`   `                    ``if``(first != NULL && ` `                       ``prev->data > curr->data)` `                    ``{` `                          ``// If this is second violation, ` `                           ``// mark this node as last ` `                        ``last = curr;` `                    ``}` `                    ``prev = curr;` `                  `  `                      ``/* Revert the changes made in the` `                    ``'if' part to restore the  ` `                    ``original tree i.e., fix the` `                    ``right child of predecessor*/` `                    ``pre->right = NULL;` `                    ``curr = curr->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( ``struct` `node* root ) ` `{ ` `    ``// Initialize pointers needed` `    ``// for correctBSTUtil() ` `    ``struct` `node* first =NULL ;` `    ``struct` `node* last = NULL ;` `    ``struct` `node* prev =NULL ;`   `    ``// Set the poiters to find out two nodes ` `    ``MorrisTraversal( root ,first ,last , prev); `   `    ``// Fix (or correct) the tree ` `    ``swap( &(first->data), &(last->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 Code */` `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; ` `} ` `// This code is contributed by ` `// Sagara Jangra and Naresh Saharan `

 `// 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, last, prev; ` `      `  `    ``// This function does inorder traversal ` `    ``// Using Morris Traversal to find out the two` `      ``// swapped nodes. ` `    ``void` `MorrisTraversal( Node root) ` `    ``{ ` `       ``// current node` `        ``Node curr = root;` `        ``Node pre = ``null``;` `        ``while``(curr != ``null``)` `        ``{` `            ``if``(curr.left==``null``)` `            ``{` `              `  `              ``// If this node is smaller than ` `              ``// the previous node, it's  ` `              ``// violating the BST rule. ` `                `  `                ``if``(first == ``null` `&& prev != ``null` `&& ` `                   ``prev.data > curr.data)` `                ``{` `                      ``// If this is first violation, ` `                    ``// mark these two nodes as ` `                    ``// 'first' and 'last'` `                    ``first = prev;` `                    ``last = curr;` `                ``}` `                `  `                ``if``(first != ``null` `&& ` `                   ``prev.data > curr.data)` `                ``{` `                  ``// If this is second violation,          ` `                  ``// mark this node as last` `                    ``last = curr;` `                ``}` `                ``prev = curr;` `                `  `                ``curr = curr.right;` `            ``}` `            ``else` `            ``{` `                  ``/* Find the inorder predecessor of current */` `                ``pre = curr.left; ` `                ``while``(pre.right!=``null` `&& ` `                      ``pre.right!=curr)` `                ``{` `                    ``pre = pre.right;` `                ``}` `                `  `                ``// Make current as right child of` `                ``// its inorder predecessor */` `                ``if``(pre.right==``null``)` `                ``{` `                    ``pre.right = curr;` `                    ``curr = curr.left;` `                ``}` `                ``else` `                ``{` `                    ``// If this node is smaller than ` `                    ``// the previous node, it's  ` `                    ``// violating the BST rule. ` `                    ``if``(first == ``null` `&& prev!=``null` `&& ` `                       ``prev.data > curr.data)` `                    ``{` `                          ``// If this is first violation, ` `                        ``// mark these two nodes as ` `                        ``// 'first' and 'last' ` `                        ``first = prev;` `                        ``last = curr;` `                    ``}`   `                    ``if``(first != ``null` `&& ` `                       ``prev.data > curr.data)` `                    ``{` `                          ``// If this is second violation, ` `                           ``// mark this node as last ` `                        ``last = curr;` `                    ``}` `                    ``prev = curr;` `                  `  `                      ``/* Revert the changes made in the` `                    ``'if' part to restore the  ` `                    ``original tree i.e., fix the ` `                    ``right child of predecessor*/` `                    ``pre.right = ``null``;` `                    ``curr = curr.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 = last = prev = ``null``; ` `  `  `        ``// Set the poiters to find out  ` `        ``// two nodes ` `        ``MorrisTraversal( root ); ` `  `  `        ``// Fix (or correct) the tree ` `          ``int` `temp = first.data; ` `        ``first.data = last.data; ` `          ``last.data = temp;  ` `    ``} ` `  `  `  `  `       ``/* 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 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``); ` `  `  `        ``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 ` `// Naresh Saharan and Sagara Jangra`

 `// C# program to correct the BST  ` `// if two nodes are swapped ` `using` `System;` `public`     ` ``class` `Node { ` `  `  `    ``public`     ` ``int` `data; ` `   ``public`     `  ``Node left, right; ` `  `  `    ``public` `Node(``int` `d) { ` `        ``data = d; ` `        ``left = right = ``null``; ` `    ``} ` `} `   `public` `class` `GFG {` `  `  `       ``Node first, last, prev; ` `      `  `    ``// This function does inorder traversal ` `    ``// Using Morris Traversal to find out the two` `      ``// swapped nodes. ` `    ``void` `MorrisTraversal( Node root) ` `    ``{ ` `       ``// current node` `        ``Node curr = root;` `        ``Node pre = ``null``;` `        ``while``(curr != ``null``)` `        ``{` `            ``if``(curr.left==``null``)` `            ``{` `              `  `              ``// If this node is smaller than ` `              ``// the previous node, it's  ` `              ``// violating the BST rule. ` `                `  `                ``if``(first == ``null` `&& prev != ``null` `&& ` `                   ``prev.data > curr.data)` `                ``{` `                      ``// If this is first violation, ` `                    ``// mark these two nodes as ` `                    ``// 'first' and 'last'` `                    ``first = prev;` `                    ``last = curr;` `                ``}` `                `  `                ``if``(first != ``null` `&& ` `                   ``prev.data > curr.data)` `                ``{` `                  ``// If this is second violation,          ` `                  ``// mark this node as last` `                    ``last = curr;` `                ``}` `                ``prev = curr;` `                `  `                ``curr = curr.right;` `            ``}` `            ``else` `            ``{` `                  ``/* Find the inorder predecessor of current */` `                ``pre = curr.left; ` `                ``while``(pre.right!=``null` `&& ` `                      ``pre.right!=curr)` `                ``{` `                    ``pre = pre.right;` `                ``}` `                `  `                ``// Make current as right child of` `                ``// its inorder predecessor */` `                ``if``(pre.right==``null``)` `                ``{` `                    ``pre.right = curr;` `                    ``curr = curr.left;` `                ``}` `                ``else` `                ``{` `                    ``// If this node is smaller than ` `                    ``// the previous node, it's  ` `                    ``// violating the BST rule. ` `                    ``if``(first == ``null` `&& prev!=``null` `&& ` `                       ``prev.data > curr.data)` `                    ``{` `                          ``// If this is first violation, ` `                        ``// mark these two nodes as ` `                        ``// 'first' and 'last' ` `                        ``first = prev;` `                        ``last = curr;` `                    ``}`   `                    ``if``(first != ``null` `&& ` `                       ``prev.data > curr.data)` `                    ``{` `                          ``// If this is second violation, ` `                           ``// mark this node as last ` `                        ``last = curr;` `                    ``}` `                    ``prev = curr;` `                  `  `                      ``/* Revert the changes made in the` `                    ``'if' part to restore the  ` `                    ``original tree i.e., fix the ` `                    ``right child of predecessor*/` `                    ``pre.right = ``null``;` `                    ``curr = curr.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 = last = prev = ``null``; ` `  `  `        ``// Set the poiters to find out  ` `        ``// two nodes ` `        ``MorrisTraversal( root ); ` `  `  `        ``// Fix (or correct) the tree ` `          ``int` `temp = first.data; ` `        ``first.data = last.data; ` `          ``last.data = temp;  ` `    ``} ` `  `  `  `  `       ``/* 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"``); ` `        ``GFG tree = ``new` `GFG(); ` `        ``tree.printInorder(root); ` `  `  `        ``tree.correctBST(root); ` `  `  `        ``Console.WriteLine(``"\nInorder Traversal"``+ ` `                          ``" of the fixed tree"``); ` `        ``tree.printInorder(root); ` `    ``} ` `} ` `// This code contributed by gauravrajput1`

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)
Auxiliary Space: O(1)

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