PAT 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

题目大意：给出一个序列， 判断其是否是一棵二叉搜索树的前序遍历， 如果是则输出其后序遍历
思路： 按照前序遍历的特性去建立一棵树
　　　　此外，二叉搜索树的中序遍历得到的序列是一个有序的序列；
　　　　那么，我们通过给出的序列建立一棵树， 再中序遍历这棵树， 如果得到的结果是有序的，则给出的序列是一颗搜索二叉树的前序遍历；

 1 #include<iostream>
 2 #include<vector>
 3 #include<queue>
 4 using namespace std;
 5 struct node{
 6   int val;
 7   node *left, *right;
 8 };
 9 
10 node *root=NULL;
11 bool flag=true, f=true; //false for mirror
12 vector<int> in, post, pre;
13 void dfs(node* &root, int l, int r){
14   if(l>r) return;
15   if(root==NULL){
16     root = new node;
17     root->val = pre[l];
18     root->left=root->right=NULL;
19   }
20   int i=l;
21   //找左右子树的分界点
22   if(flag) while(i<=r && pre[i]<pre[l] || i==l) i++;
23   else while(i<=r && pre[i]>=pre[l]) i++; 
24   dfs(root->left, l+1, i-1);
25   dfs(root->right, i, r);
26 }
27 
28  
29 void postOrder(node* root){
30     if(root==NULL) return;
31     if(root->left)    postOrder(root->left);
32     if(root->right)    postOrder(root->right);
33     post.push_back(root->val);
34 }
35 
36 void inOrder(node* root){
37     if(root==NULL) return;
38     if(root->left) inOrder(root->left);
39     in.push_back(root->val);
40     if(root->right) inOrder(root->right);
41 }
42 
43 int main(){
44   int n, i;
45   scanf("%d", &n);
46   pre.resize(n);
47   for(i=0; i<n; i++) scanf("%d", &pre[i]);
48   //如果pre[1]>=pre[0]则这是镜像的前序遍历
49   if(pre[0]<=pre[1]) flag=false;
50   dfs(root, 0, n-1);
51   inOrder(root);
52   for(i=0; i<n-1; i++){//判断中序遍历是否有序
53       if(in[i]>in[i+1] && flag) f=false; 
54       if(in[i]<in[i+1] && !flag) f=false; 
55   }
56   if(!f) printf("NO");
57   else{
58     postOrder(root);
59     printf("YES\n");
60     for(i=0; i<n; i++){
61         if(i!=0) printf(" ");
62         printf("%d", post[i]);
63     }
64   }
65   return 0;
66 }

 

转载于:https://www.cnblogs.com/mr-stn/p/9229552.html