PAT 甲级 1043 Is It a Binary Search Tree

https://pintia.cn/problem-sets/994805342720868352/problems/994805440976633856

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

 

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
vector<int> pre, post;
bool isMirror;

void solve(int root, int point) {
    if(root > point) return ;
    int i = root + 1, j = point;
    if(!isMirror) {
        while(i <= point && pre[root] > pre[i]) i ++;
        while(j > root && pre[root] <= pre[i]) j --;
    } else {
        while(i <= point && pre[root] <= pre[i]) i ++;
        while(j > root && pre[root] > pre[j]) j --;
    }

    if(i - j != 1) return ;
    solve(root + 1, j);
    solve(i, point);
    post.push_back(pre[root]);
}

int main() {
    scanf("%d", &N);
    pre.resize(N);
    for(int i = 0; i < N; i ++)
        scanf("%d", &pre[i]);
    solve(0, N - 1);
    if(post.size() != N) {
        isMirror = true;
        post.clear();
        solve(0, N - 1);
    }

    if(post.size() == N) {
        printf("YES\n");
z        for(int i = 0; i < N; i ++) {
            printf("%d", post[i]);
            printf("%s", i != N - 1 ? " " : "\n");
        }
    } else printf("NO\n");
    return 0;
}

　　先假设不是镜面的树 按照二叉搜索树的性质进行查找 并且按照后序遍历存起来 如果后序遍历存起来之后不是 N 个 则 isMirror = true 反着再查一遍 如果反过来查一遍之后 post.size() = N 的话说明是镜面的 如果还不是的话就是 NO 然后输出

期末被高数折磨的死去活来 好多天没摸键盘了 手好生 哭唧唧 今天想学学建树 期末期末快过去吧！！！