PAT 1043 Is It a Binary Search Tree（二叉查找树的前序遍历转化成后序遍历）

1043 Is It a Binary Search Tree

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

总结：这道题目基本不会写，可能是由于没有系统的看过算法数吧，不过也算是学到了东西，补上了自己没接触过的东西（二叉查找树的前序遍历转化成后续遍历）

#include 
#include 
using namespace std;

bool isMirror;
vector pre,post;
void getpost(int root,int tail){
    int i=root+1,j=tail;
    if(!isMirror){
        while(i<=tail && pre[i]root && pre[j]>=pre[root])  j--;
    }
    else{
        while(i<=tail && pre[i]>=pre[root]) i++;
        while(j>root && pre[j]> n;
    pre.resize(n);
    for(int i=0;i

 好好学习，天天向上！

我要考研！