1043 Is It a Binary Search Tree (25 分)——甲级（二叉搜索树）

A 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

题目大意：给定一个序列，判断它是不是一颗BST的前序或镜像前序。如果是，输出YES以及后序或镜像后序；否则，输出NO。
思路：先假设是一颗BST，通过序列来建立这棵树。然后遍历这棵树即可。

代码如下:

#include <iostream>
#include <vector>
using namespace std;
typedef struct node* bst;
struct node
{
    int data;
    bst left;
    bst right;
};
vector<int>v, pre, mir;
bst Insert(bst t, int x)//建树
{
    if(!t)
    {
        t = new node;
        t->data = x;
        t->left = t->right = NULL;
    }
    else if(x >= t->data)
        t->right = Insert(t->right, x);
    else
        t->left = Insert(t->left, x);
    return t;
}
void preorder(bst t)//前序
{
    if(!t) return ;
    pre.push_back(t->data);
    preorder(t->left);
    preorder(t->right);
}
void mirpre(bst t)//镜像前序
{
    if(!t) return ;
    mir.push_back(t->data);
    mirpre(t->right);
    mirpre(t->left);
}
void postorder(bst t)//后序
{
    if(!t) return ;
    postorder(t->left);
    postorder(t->right);
    v.push_back(t->data);
}
void mirpost(bst t)//镜像后序
{
    if(!t) return ;
    mirpost(t->right);
    mirpost(t->left);
    v.push_back(t->data);
}
int main()
{
    int n;
    cin >> n;
    bst t = NULL;
    int i;
    for(i=0;i<n;i++)
    {
        int x;
        cin >> x;
        v.push_back(x);
        t = Insert(t, x);
    }
    preorder(t);
    mirpre(t);
    if(v == pre)
    {
        cout << "YES" <<endl;
        v.clear();
        postorder(t);
        cout << v[0];
        for(i=1; i<v.size(); i++)
            cout << " " << v[i];
    }
    else if(v == mir)
    {
        cout << "YES" <<endl;
        v.clear();
        mirpost(t);
        cout << v[0];
        for(i=1; i<v.size(); i++)
            cout << " " << v[i];
    }
    else cout << "NO";
    cout << endl;
    return 0;
}

总结

1. 今天才知道vector是可以直接判断是否相等的。。
2. 关于BST的题，有时候题目会说左子树小于等于根，有时候题目会说右子树大于等于根。要注意等于是在哪边。