输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历的结果。如果是则输出Yes,否则输出No。假设输入的数组的任意两个数字都互不相同。

// ConsoleApplication2.cpp : 定义控制台应用程序的入口点。

//

#include "stdafx.h"
#include "stdafx.h"
#include<iostream>
#include<vector>
#include<algorithm>
#include<numeric>
#include<list>
#include<iterator>
#include<queue>
#include<stack>
using namespace std;

struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
	val(x), left(NULL), right(NULL) {
	}
};

class Solution {
public:
	bool VerifySquenceOfBST(vector<int> sequence) {
		vector<int> leftSubTree;//存入左孩子
		vector<int> rightSubTree;//存入右孩子
		int root = 0; //存放根节点的值
		int flag = 0;//根据flag的值决定结果存入右孩子还是左孩子
		bool result = true;
		if (sequence.empty()) return false;//如果sequence为空的时候返回false
			root = *(sequence.end() - 1);
			for (auto it = sequence.begin(); it != sequence.end()-1; it++) //划分左孩子和右孩子
			{
				if (*it < root&&flag == 0)
					leftSubTree.push_back(*it);
				else if (*it > root)
				{
					flag = 1;
					rightSubTree.push_back(*it);
				}
				else
				{
					rightSubTree.push_back(*it);
				}
			}
			//根据左孩子和右孩子判断结果，左孩子的所有值都小于根节点，右孩子的所有值都大于根节点
			//这里左孩子不用判断，因为上述存储中已经保证

			for (auto it = rightSubTree.begin(); it != rightSubTree.end(); ++it)
			{
				if (*it < root)
				{
					result = false;
					break;
				}
			}
			if (result == false) //第一次检查错误，不在递归，范围false
				return false;

			if (!rightSubTree.empty()&&!leftSubTree.empty()) //进行下次递归时要判断左子树和右子树是否为空，为空的时候不需要进行递归
			   result = VerifySquenceOfBST(rightSubTree)&&VerifySquenceOfBST(leftSubTree);
			else if(!rightSubTree.empty() && leftSubTree.empty())
				result = VerifySquenceOfBST(rightSubTree) ;
			else if (rightSubTree.empty() && !leftSubTree.empty())
				result = VerifySquenceOfBST(leftSubTree);
			else
				result = true;
			return result;
	}
};

int main()
{

	Solution so;

	vector<int> tree1 = { 5,7, 6, 9, 11, 10, 8 };
	vector<int> tree2 = { 7,4,6,5 };

	bool result1 = so.VerifySquenceOfBST(tree1);
	cout << "tree1结果是：" ;
	cout << result1 << endl;
	cout << endl;

	bool result2 = so.VerifySquenceOfBST(tree2);
	cout << "tree2结果是：";
	cout << result2 << endl;
	cout << endl;
	return 0;
}