似乎寫複雜了，列舉所有的結果

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
    struct node {
        int l, r;
    };
    vector<int> sta;
    vector
 
<TreeNode*> q, ans;
    vector<node> p;
    void inorderTraversal(TreeNode* &root) {
        vector<TreeNode*> sta;
        int num = 1;
        sta.push_back(root);
        while (sta.back()->left) sta.push_back(sta.back()->left);
        while (!sta.empty()) {
            TreeNode* p = sta.back(); sta.pop_back();
            p->val = num++;
            if
 
 (p->right) {
                sta.push_back(p->right);
                while (sta.back()->left) sta.push_back(sta.back()->left);
            }
        }
    }
    void dfs(int next, int &n, int &ptr) {
        if (next > n) {
            ptr = 0;
            for (auto &x : sta) p[x].r = 0
 
;
            for (int i = 0; i < n; ++i)
                q[i] = new TreeNode(0);
            for (int i = 0; i < n; ++i)
                q[i]->left = p[i + 1].l ? q[p[i + 1].l - 1] : NULL, q[i]->right = p[i + 1].r ? q[p[i + 1].r - 1] : NULL;
            inorderTraversal(q[0]);
            ans.push_back(q[0]);
            return;
        }
        if (sta.empty()) {
            ptr = next;
            sta.push_back(next);
            dfs(next + 1, n, p[next].l);
            sta.pop_back();
        }
        else {
            ptr = next;
            sta.push_back(next);
            dfs(next + 1, n, p[next].l);
            sta.pop_back();

            ptr = 0;
            int k = sta.back();
            sta.pop_back();
            dfs(next, n, p[k].r);
            sta.push_back(k);
        }
    }
    vector<TreeNode*> generateTrees(int n) {
        if (!n) return ans;
        p.resize(n + 1), q.resize(n);
        sta.push_back(1), dfs(2, n, p[1].l);
        return ans;
    }
};