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

1043. Is It a Binary Search Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

78 6 5 7 10 8 11

Sample Output 1:

YES5 7 6 8 11 10 8

Sample Input 2:

78 10 11 8 6 7 5

Sample Output 2:

YES11 8 10 7 5 6 8

Sample Input 3:

78 6 8 5 10 9 11

Sample Output 3:

NO

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这道题只要求判断是否为BST或镜像BST，如果是的话按照Postorder输出整个树，因此实际上不用真的建一个二叉树。

只要根据BST的性质在一列数中区分出顶点，左子树，右子树，然而递归判断就可以判断出这个数是否为BST。

同样的，也可以按照这样的方法后序输出整个数。代码如下： 

#include <iostream>using namespace std;int ori[1000];bool isMirror;bool judge(int begin,int end){int i,j,root=ori[begin];if(begin>end)return true;for(i=begin+1;i<=end;i++){if(ori[i]>=root&&!isMirror)break;if(ori[i]<root&&isMirror)break;}for(j=i;j<=end;j++){if(ori[j]<root&&!isMirror)return false;if(ori[j]>=root&&isMirror)return false;}int leftbegin=begin+1,leftend=i-1,rightbegin=i,rightend=end;if(!judge(leftbegin,leftend))return false;if(!judge(rightbegin,rightend))return false;return true;}void print(int begin,int end,int level){if(begin>end)return;int root=ori[begin],i;for(i=begin+1;i<=end;i++){if(ori[i]>=root&&!isMirror)break;if(ori[i]<root&&isMirror)break;}int leftbegin=begin+1,leftend=i-1,rightbegin=i,rightend=end;print(leftbegin,leftend,level+1);print(rightbegin,rightend,level+1);cout<<root;if(level>1)cout<<" ";return;}int main(void){int K,i;cin>>K;for(i=0;i<K;i++)cin>>ori[i];if(ori[1]<ori[0])isMirror=false;elseisMirror=true;if(judge(0,K-1)){cout<<"YES"<<endl;print(0,K-1,1);}elsecout<<"NO";}