PAT1043 Is It a Binary Search Tree （25）（二叉查找树BST）

类型

二叉搜索树, 先序后序转换

题目

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

分析题目

输入一个先序的树，判断是否为二叉搜索树，或者是否为二叉搜索树的镜像树。

变量	意义
root	根节点
root + 1	左子树第一个结点
tail	右子树最后一个结点
i	右子树第一个结点
j	左子树最后一个结点
isMirror	是否为镜像二叉搜索树的flag

利用二叉搜索树的性质：左子树下的所有节点比根节点要小，右子树的所有节点节点比根节点大或者相等

• 先通过二叉树性质判断是否为先序的二叉搜索树，如果是的话，getpost获得的后序树节点数等于N，否则小于N。
• 若不为先序的二叉搜索树，则将isMirror设置为true（先假设它是镜像的二叉搜索树），通过二叉树性质（此时左子树节点比根节点大或相等，右子树节点比根节点小）判断是否为镜像二叉搜索树。

代码

#include 
#include 
using namespace std;
vector pre, post;
int isMirror = 0;
void getpost(int root, int tail)
{
    if(root > tail)
        return;
    int i = root+1;
    int 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> pre[i];
    }
    getpost(0, N-1);
    if(post.size()!=N)
    {
        isMirror = 1;
        post.clear();
        getpost(0, N-1);
    }
    if(post.size() == N)
    {
        cout << "YES" << endl << post[0];
        for(int i=1; i

总结

学习了二叉搜索树的性质，对先序和后序的转化复习了一下，学(chao)习还是来自于柳神的代码