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A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
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 11Sample Output 1:
YES 5 7 6 8 11 10 8Sample Input 2:
7 8 10 11 8 6 7 5Sample Output 2:
YES 11 8 10 7 5 6 8Sample Input 3:
7 8 6 8 5 10 9 11Sample Output 3:
NO
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<queue> 6 #include<vector> 7 #include<cmath> 8 #include<string> 9 #include<map> 10 #include<set> 11 using namespace std; 12 struct node 13 { 14 int v; 15 node *l,*r; 16 node() 17 { 18 l=r=NULL; 19 } 20 }; 21 bool Buildtree(node *&h,int *line,int n) 22 { 23 if(n==0) 24 { 25 return true; 26 } 27 int i; 28 h=new node(); 29 h->v=line[0]; 30 i=1; 31 while(i<n&&line[i]<line[0]) 32 { 33 i++; 34 } 35 int j=i; 36 while(j<n&&line[j]>=line[0]){ 37 j++; 38 } 39 if(j!=n){ 40 return false; 41 } 42 return Buildtree(h->l,line+1,i-1)&&Buildtree(h->r,line+i,n-i); 43 } 44 45 bool Buildtree1(node *&h,int *line,int n) 46 { 47 if(n==0) 48 { 49 return true; 50 } 51 int i; 52 h=new node(); 53 h->v=line[0]; 54 i=1; 55 while(i<n&&line[i]>=line[0]) 56 { 57 i++; 58 } 59 int j=i; 60 while(j<n&&line[j]<line[0]){ 61 j++; 62 } 63 if(j!=n){ 64 return false; 65 } 66 return Buildtree1(h->l,line+1,i-1)&&Buildtree1(h->r,line+i,n-i);//黏贴复制害死人 67 } 68 void Postorder(node *&h) 69 { 70 if(h) 71 { 72 Postorder(h->l); 73 Postorder(h->r); 74 printf("%d ",h->v); 75 delete []h; 76 } 77 } 78 int line[1005]; 79 int main() 80 { 81 //freopen("D:\\INPUT.txt","r",stdin); 82 int n; 83 while(scanf("%d",&n)!=EOF) 84 { 85 node *h; 86 int i; 87 for(i=0; i<n; i++) 88 { 89 scanf("%d",&line[i]); 90 } 91 if(n==1){ 92 printf("YES\n"); 93 printf("%d\n",line[0]); 94 continue; 95 } 96 if(line[0]>line[1]&&Buildtree(h,line,n))//BST 97 { 98 printf("YES\n"); 99 Postorder(h->l); 100 Postorder(h->r); 101 printf("%d\n",h->v); 102 continue; 103 } 104 if(line[0]<=line[1]&&Buildtree1(h,line,n)) 105 { 106 printf("YES\n"); 107 Postorder(h->l); 108 Postorder(h->r); 109 printf("%d\n",h->v); 110 continue; 111 } 112 printf("NO\n"); 113 } 114 return 0; 115 }
pat1043. Is It a Binary Search Tree (25)
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原文地址：http://www.cnblogs.com/Deribs4/p/4770374.html
