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 auxiilary array elements back to the BST, keeping the structure of the BST same. Time complexity of this method is O(nLogn) and 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 ( 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 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 current node value is smaller than previous node value, we update the first with the previous node & middle with the current node. When we find the second point where current node value is smaller than previous node value, we update the last with the current node. In case #2, we will never find the second point. So, 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 C implementation of the given code.

// 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; }

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.

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