PAT Advanced 1043 Is It a Binary Search Tree (25分)

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 (≤). 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

这道题目考察了二叉排序树。二叉排序树一个性质，已知先序就可以求后序。

1.我们可以直接使用sort进行排序，然后根据先序中序求后序。

2.我们可以用i，j进行把二叉树划分2个子树，进行递归求解。（参考柳神）

#include <iostream>
#include <vector>
using namespace std;
int N;
bool mirror = false;
vector<int> pre, post;
void getPost(int root, int tail){
    if(root > tail) return ;
    int i = root + 1, j = tail;
    if(!mirror) {
        while(i <= tail && pre[i] < pre[root]) i++;//右子树最小的
        while(j > root && pre[j] >= pre[root]) j--;//左子树最大的
    }else {
        while(i <= tail && pre[i] >= pre[root]) i++;//右子树最小的
        while(j > root && pre[j] < pre[root]) j--;//左子树最大的
    }
    if(i - j != 1) return ; // 此时相差必为1, 否则不符合
    getPost(root + 1, j);
    getPost(i, tail);
    post.push_back(pre[root]);
}
int main() {
    cin >> N;
    pre.resize(N);
    for(int i = 0; i < N; i++)
        cin >> pre[i];
    getPost(0, N-1);
    if(post.size() != N) {
        post.clear();
        mirror = true;
        getPost(0, N-1);
    }
    if(post.size() != N) cout << "NO";
    else {
        cout << "YES" << endl;
        for(int i = 0; i < N; i++)
            if(i != N-1) cout << post[i] <<" ";
            else cout << post[i];
    }
    system("pause");
    return 0;
}

原文：https://www.cnblogs.com/littlepage/p/12230017.html