JZ33 二叉搜索树的后序遍历序列

class Solution {
public:
    bool VerifySquenceOfBST(vector<int> s) {
        int n = s.size();
        if (n==0)return 0;
        int l = 0, r = n - 1;
        return check(s, l, r);
    }
    bool check (vector<int>&sq, int l, int r) {
        if (l >= r) return 1;
        int root = sq[r];
        int j = r - 1;
        while (j>= l && sq[j] > root) j --;
        for (int i = l; i <= j; i ++) {
            if (sq[i] > root) return 0;
        }
        return check(sq, l, j) && check(sq, j + 1, r - 1);
    }
};

https://www.nowcoder.com/practice/a861533d45854474ac791d90e447bafd?tpId=13&tqId=23289&ru=%2Fpractice%2F508378c0823c423baa723ce448cbfd0c&qru=%2Fta%2Fcoding-interviews%2Fquestion-ranking&sourceUrl= 问题结构:性质题

我们可以确定最后一个r一定是根节点，然后我们可以划分左子树和右子树判断就行了