1043 Is It a Binary Search Tree (25 分)

时间：2019-07-09 14:00:03 阅读：135 评论：0 收藏：0 [点我收藏+]

标签：numbers -o with follow clu false seq problem span

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

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

判断是否是是二叉搜索树或镜像二叉搜索树。

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 int n, an[1005];
  5 
  6 struct Node
  7 {
  8     int val;
  9     Node *left;
 10     Node *right;
 11 }*head;
 12 int pre[1005];
 13 int pos = 0;
 14 
 15 void insert(Node* &root, int x){
 16     if(root == NULL){
 17         root = new Node;
 18         root->val = x;
 19         root->left = NULL;
 20         root->right = NULL;
 21     }else{
 22         if(x < root->val){
 23             insert(root->left, x);
 24         }else{
 25             insert(root->right, x);
 26         }
 27     }
 28 }
 29 
 30 void prefind(Node * &root){
 31     if(root != NULL){
 32         pre[pos++] = root->val;
 33         prefind(root->left);
 34         prefind(root->right);
 35     }
 36 }
 37 
 38 void premirror(Node * &root){
 39     if(root != NULL){
 40         pre[pos++] = root->val;
 41         premirror(root->right);
 42         premirror(root->left);
 43     }
 44 }
 45 
 46 void behind(Node * &root){
 47     if(root != NULL){
 48         behind(root->left);
 49         behind(root->right);
 50         pre[pos++] = root->val;
 51     }
 52 }
 53 
 54 void mirror(Node * &root){
 55     if(root != NULL){
 56         mirror(root->right);
 57         mirror(root->left);
 58         pre[pos++] = root->val;
 59     }
 60 }
 61 
 62 
 63 int main(){
 64     cin >> n;
 65     for(int i = 0; i < n; i++){
 66         cin >> an[i];
 67         insert(head, an[i]);
 68     }
 69     bool flag = true;
 70     prefind(head);
 71     for(int i = 0; i < n; i++){
 72         if(pre[i] != an[i]){
 73             flag = false;
 74             break;
 75         }
 76     }
 77     memset(pre,0,sizeof(pre));
 78     if(flag){
 79         pos = 0;
 80         behind(head);
 81         cout <<"YES"<<endl;
 82         for(int i = 0; i < n; i++)
 83             printf("%d%c", pre[i], i==n-1?‘\n‘:‘ ‘);
 84         return 0;
 85     }
 86     pos = 0;
 87     flag = true;
 88     memset(pre,0,sizeof(pre));
 89     premirror(head);
 90     for(int i = 0; i < n; i++){
 91         if(pre[i] != an[i]){
 92             flag = false;
 93             break;
 94         }
 95     }
 96     memset(pre,0,sizeof(pre));
 97     if(flag){
 98         pos = 0;
 99         mirror(head);
100         cout <<"YES"<<endl;
101         for(int i = 0; i < n; i++)
102             printf("%d%c", pre[i], i==n-1?‘\n‘:‘ ‘);
103         return 0;
104     }
105     cout <<"NO"<<endl;
106     return 0;
107 }

1043 Is It a Binary Search Tree (25 分)

标签：numbers -o with follow clu false seq problem span

原文地址：https://www.cnblogs.com/zllwxm123/p/11156565.html

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