剑指 Offer 33. 二叉搜索树的后序遍历序列

class Solution {
public:
    bool buildTree(vector<int> &postorder, vector<int> &inorder, int postL, int postR, int inL, int inR) {
        if(postR == postL && inR == inL) return true;
        if(postR - postL != inR - inL) return false;
        int cnt = 0;
        while(inL + cnt < inR && inorder[inL + cnt] != postorder[postR-1]) cnt++;
        if(inL + cnt >= inR) return false;
        bool left_flag = buildTree(postorder, inorder, postL, postL+cnt, inL, inL+cnt);
        bool right_flag = buildTree(postorder, inorder, postL+cnt, postR-1, inL+cnt+1, inR);
        return left_flag && right_flag;
    }
    bool verifyPostorder(vector<int>& postorder) {
        vector<int> inorder(postorder.begin(), postorder.end());
        sort(inorder.begin(), inorder.end());  // 对后序遍历结果排序可以得到中序遍历
        return buildTree(postorder, inorder, 0, postorder.size(), 0, inorder.size());  // 根据中后序还原BST看是否出错
    }
};