Two nodes of a BST are swapped, correct the BST[转载]
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
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
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 the implementation of the given code.
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 }