46.二叉搜索树的后序遍历序列

class Solution {
public:
    vector<int> seq;
    
    bool verifySequenceOfBST(vector<int> sequence) {
        seq = sequence;
        return dfs(0, seq.size() - 1);
    }
    bool dfs(int l, int r) {
        if (l >= r)  return true; //空树
        int root = seq[r];
        int k = l;
        while(k < r && seq[k] < root) k ++;  //找到根节点的左子树节点
        for(int i = k; i < r; i ++) {  //判断一下右子树里面的所有点都比根节点大,满足，则合法
            if (seq[i] < root)  //不满足，返回false
                return false;
        }
        return dfs(l, k - 1) && dfs(k, r-1);//递归判断左右子树

    }
};