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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 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
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <iostream> 4 #include <string.h> 5 #include <math.h> 6 #include <algorithm> 7 #include <string> 8 #include <stack> 9 #include <queue> 10 using namespace std; 11 int n; 12 struct tree{ 13 int data; 14 tree *left; 15 tree *right; 16 }; 17 18 void insert(tree *&root,int data) 19 { 20 if(root==NULL) 21 { 22 root=new tree; 23 root->data=data; 24 root->left=NULL; 25 root->right=NULL; 26 return ; 27 } 28 if(data<root->data)insert(root->left,data); 29 else insert(root->right,data); 30 } 31 32 void preorder(tree * root,vector<int> &pre) 33 { 34 if(root==NULL)return; 35 pre.push_back(root->data); 36 preorder(root->left,pre); 37 preorder(root->right,pre); 38 39 } 40 void preorderM(tree * root,vector<int> &preM) 41 { 42 if(root==NULL)return ; 43 preM.push_back(root->data); 44 preorderM(root->right,preM); 45 preorderM(root->left,preM); 46 47 } 48 void postorder(tree * root,vector<int> &post) 49 { 50 if(root==NULL)return ; 51 postorder(root->left,post); 52 postorder(root->right,post); 53 post.push_back(root->data); 54 } 55 void postorderM(tree * root,vector<int> &postM) 56 { 57 if(root==NULL)return; 58 postorderM(root->right,postM); 59 postorderM(root->left,postM); 60 postM.push_back(root->data); 61 } 62 63 vector<int> origin,pre,preM,post,postM; 64 int main(){ 65 scanf("%d",&n); 66 tree * root=NULL; 67 for(int i=0;i<n;i++) 68 { 69 int temp; 70 scanf("%d",&temp); 71 origin.push_back(temp); 72 insert(root,temp); 73 } 74 75 //树已经插入好，开始先序遍历 76 preorder(root,pre); 77 //镜像先序遍历 78 preorderM(root,preM) ; 79 80 81 postorder(root,post); 82 postorderM(root,postM); 83 if(origin==pre) 84 { 85 printf("YES\n"); 86 for(int i=0;i<post.size();i++) 87 { 88 printf("%d",post[i]); 89 if(i<post.size()-1)printf(" "); 90 } 91 }else if(origin==preM) 92 { 93 printf("YES\n"); 94 for(int i=0;i<postM.size();i++) 95 { 96 printf("%d",postM[i]); 97 if(i<postM.size()-1)printf(" "); 98 } 99 }else 100 { 101 printf("NO\n"); 102 } 103 104 105 106 return 0; 107 }
A1043. Is It a Binary Search Tree (25)
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原文地址：http://www.cnblogs.com/ligen/p/4319375.html
