Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

  1 // Two nodes in the BST's swapped, correct the BST.
  2 #include <stdio.h>
  3 #include <stdlib.h>
  4
  5 /* A binary tree node has data, pointer to left child
  6    and a pointer to right child */
  7 struct node
  8 {
  9     int data;
 10     struct node *left, *right;
 11 };
 12
 13 // A utility function to swap two integers
 14 void swap( int* a, int* b )
 15 {
 16     int t = *a;
 17     *a = *b;
 18     *b = t;
 19 }
 20
 21 /* Helper function that allocates a new node with the
 22    given data and NULL left and right pointers. */
 23 struct node* newNode(int data)
 24 {
 25     struct node* node = (struct node *)malloc(sizeof(struct node));
 26     node->data = data;
 27     node->left = NULL;
 28     node->right = NULL;
 29     return(node);
 30 }
 31
 32 // This function does inorder traversal to find out the two swapped nodes.
 33 // It sets three pointers, first, middle and last.  If the swapped nodes are
 34 // adjacent to each other, then first and middle contain the resultant nodes
 35 // Else, first and last contain the resultant nodes
 36 void correctBSTUtil( struct node* root, struct node** first,
 37                      struct node** middle, struct node** last,
 38                      struct node** prev )
 39 {
 40     if( root )
 41     {
 42         // Recur for the left subtree
 43         correctBSTUtil( root->left, first, middle, last, prev );
 44
 45         // If this node is smaller than the previous node, it's violating
 46         // the BST rule.
 47         if (*prev && root->data < (*prev)->data)
 48         {
 49             // If this is first violation, mark these two nodes as
 50             // 'first' and 'middle'
 51             if ( !*first )
 52             {
 53                 *first = *prev;
 54                 *middle = root;
 55             }
 56
 57             // If this is second violation, mark this node as last
 58             else
 59                 *last = root;
 60         }
 61
 62         // Mark this node as previous
 63         *prev = root;
 64
 65         // Recur for the right subtree
 66         correctBSTUtil( root->right, first, middle, last, prev );
 67     }
 68 }
 69
 70 // A function to fix a given BST where two nodes are swapped.  This
 71 // function uses correctBSTUtil() to find out two nodes and swaps the
 72 // nodes to fix the BST
 73 void correctBST( struct node* root )
 74 {
 75     // Initialize pointers needed for correctBSTUtil()
 76     struct node *first, *middle, *last, *prev;
 77     first = middle = last = prev = NULL;
 78
 79     // Set the poiters to find out two nodes
 80     correctBSTUtil( root, &first, &middle, &last, &prev );
 81
 82     // Fix (or correct) the tree
 83     if( first && last )
 84         swap( &(first->data), &(last->data) );
 85     else if( first && middle ) // Adjacent nodes swapped
 86         swap( &(first->data), &(middle->data) );
 87
 88     // else nodes have not been swapped, passed tree is really BST.
 89 }
 90
 91 /* A utility function to print Inoder traversal */
 92 void printInorder(struct node* node)
 93 {
 94     if (node == NULL)
 95         return;
 96     printInorder(node->left);
 97     printf("%d ", node->data);
 98     printInorder(node->right);
 99 }
100
101 /* Driver program to test above functions*/
102 int main()
103 {
104     /*   6
105         /  \
106        10    2
107       / \   / \
108      1   3 7  12
109      10 and 2 are swapped
110     */
111
112     struct node *root = newNode(6);
113     root->left        = newNode(10);
114     root->right       = newNode(2);
115     root->left->left  = newNode(1);
116     root->left->right = newNode(3);
117     root->right->right = newNode(12);
118     root->right->left = newNode(7);
119
120     printf("Inorder Traversal of the original tree \n");
121     printInorder(root);
122
123     correctBST(root);
124
125     printf("\nInorder Traversal of the fixed tree \n");
126     printInorder(root);
127
128     return 0;
129 }