Leetcode#95 Unique Binary Search Trees II

原题地址

Unique Binary Search Trees（参见这篇文章）的升级版

做题的时候我在想，这要是把每个二叉树都独立创建一份得多麻烦啊，试试能不能共用"公共部分"，试了一下，果然可以，哈哈。

trees[i][j]表示数字i到j所能组成的所有二叉树的根节点

代码：

vector<TreeNode *> generateTrees(int n) {
        if (n < 1)
            return vector<TreeNode *>(1, NULL);

        vector<vector<vector<TreeNode *> > > trees(n + 1, vector<vector<TreeNode *> >(n + 1, vector<TreeNode *>()));

        for (int len = 1; len <= n; len++) {
            for (int i = 1, j = i + len - 1; j <= n; i++, j++) {
                for (int k = i; k <= j; k++) {
                    vector<TreeNode *> lefts = k == i ? vector<TreeNode *>(1, NULL) : trees[i][k - 1];
                    vector<TreeNode *> rights = k == j ? vector<TreeNode *>(1, NULL) : trees[k + 1][j];
                    for (auto l : lefts) {
                        for (auto r : rights) {
                            TreeNode *node = new TreeNode(k);
                            node->left = l;
                            node->right = r;
                            trees[i][j].push_back(node);
                        }
                    }
                }
            }
        }
        
        return trees[1][n];
}