判断是否后序遍历

情况：

	全右子树
    全左子树
    左右子树
    右左子树（此时不是后序遍历false）

class Solution {
public:
    bool VerifySquenceOfBST(vector<int> sequence) {
        if(sequence.size() == 0)return false;
        if(sequence.size() <= 2)return true;
        int n = sequence.size();//要么全左子树，要么全右，要么左右。
        int ans = sequence[n-1];//根节点
        int k = 0;
        bool flag = false;
        for(int i = 0;i < n;++i){
            if(sequence[i] < ans)
                k = i;
            if(ans < sequence[i])
                flag = true;
            if(flag && ans > sequence[i])
                return false;//右左穿插,直接返回
        }
        vector<int> sequenceleft(sequence.begin(),sequence.begin()+k+1);
        vector<int> sequenceright(sequence.begin()+k+1,sequence.end()-1);
        bool f = false;
        if(sequenceleft.size())//空树为false，判断，此时空树应该为真所以非空再递归
            f = VerifySquenceOfBST(sequenceleft);
        if(sequenceright.size())
            f = f && VerifySquenceOfBST(sequenceright);
        return f;
    }
};