PAT 1043. Is It a Binary Search Tree (25)

1043. Is It a Binary Search Tree (25)

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

  • The left subtree of a node contains only nodes with keys less than the node's key.
  • The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
  • Both the left and right subtrees must also be binary search trees.

If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line "YES" if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or "NO" if not. Then if the answer is "YES", print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1: 
7
8 6 5 7 10 8 11
Sample Output 1: 
YES
5 7 6 8 11 10 8
Sample Input 2: 
7
8 10 11 8 6 7 5
Sample Output 2: 
YES
11 8 10 7 5 6 8
Sample Input 3: 
7
8 6 8 5 10 9 11
Sample Output 3: 
NO

本题先判断是否的mirror BST，然后分别构造出相应的搜索树，分别对其进行中序和后序遍历。如果其为BST的话，那么其中序遍历的序列因为递增序列，mirror BST为递减序列，以此来判断
 1 #include <iostream>
 2 #include <vector>
 3 
 4 using namespace std;
 5 
 6 struct Node
 7 {
 8     int value;
 9     Node* left;
10     Node* right;
11 };
12 
13 int sequence[1000];
14 vector<int> inSequence;
15 vector<int> postSequence;
16 bool isMirror = false;
17 
18 Node* ConstrutTree(int left, int right)
19 {
20     if (left == right)
21         return nullptr;
22 
23     Node* root = new Node;
24     root->value = sequence[left];
25     int index;
26     if (isMirror)    //mirror BST
27     {
28         for (index = left + 1; index < right; index++)
29             if (sequence[index] < sequence[left])
30                 break;
31     }
32     else    //BST
33     {
34         for (index = left + 1; index < right; index++)
35             if (sequence[index] >= sequence[left])
36                 break;
37     }
38     root->left = ConstrutTree(left + 1, index);
39     root->right = ConstrutTree(index, right);
40 
41     return root;
42 }
43 
44 void InOrderTraversal(Node* root)
45 {
46     if (root == nullptr)
47         return;
48     InOrderTraversal(root->left);
49     inSequence.push_back(root->value);
50     InOrderTraversal(root->right);
51 }
52 
53 void PostOrderTraversal(Node* root)
54 {
55     if (root == nullptr)
56         return;
57     PostOrderTraversal(root->left);
58     PostOrderTraversal(root->right);
59     postSequence.push_back(root->value);
60 }
61 
62 int main()
63 {
64     int nodeNum;
65     cin >> nodeNum;
66     for (int i = 0; i < nodeNum; i++)
67         cin >> sequence[i];
68 
69     //judge if it is the mirror of BST
70     if (nodeNum>1 && sequence[1] > sequence[0])
71         isMirror = true;
72 
73     Node* root = ConstrutTree(0, nodeNum);
74     InOrderTraversal(root);
75     PostOrderTraversal(root);
76     bool flag1 = true, flag2 = true;
77     for (int i = 1; i < inSequence.size(); i++)
78     {
79         if (inSequence[i] < inSequence[i - 1])
80             flag1 = false;
81         if (inSequence[i] > inSequence[i - 1])
82             flag2 = false;
83         if (!flag1&&!flag2)
84             break;
85     }
86         
87             
88     if (flag1 || flag2)
89     {
90         cout << "YES" << endl;
91         for (int j = 0; j < postSequence.size() - 1; j++)
92             cout << postSequence[j] << " ";
93         cout << postSequence.back();
94     }
95     else
96         cout << "NO";        
97 }

 



posted @ 2015-08-11 14:16  JackWang822  阅读(226)  评论(0)  编辑  收藏  举报