LeetCode: Unique Binary Search Trees II

贴下自己写的不用指针的代码，更加容易懂吧

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode *> dfs(int start, int end) {
        vector<TreeNode *> subtree;
        if (start > end) {
            subtree.push_back(NULL);
            return subtree;
        }
        for (int i = start; i <= end; i++) {
            vector<TreeNode *> left = dfs(start, i-1);
            vector<TreeNode *> right = dfs(i+1, end);
            for (int j = 0; j < left.size(); j++) {
                for (int k = 0; k < right.size(); k++) {
                    TreeNode *tmp = new TreeNode(i);
                    tmp->left = left[j];
                    tmp->right = right[k];
                    subtree.push_back(tmp);
                }
            }
        }
        return subtree;
    }
    vector<TreeNode *> generateTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return dfs(1, n);
    }
};

 C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<TreeNode> GenerateTrees(int n) {
        return dfs(1, n);
    }
    public List<TreeNode> dfs(int start, int end) {
        List<TreeNode> subTree = new List<TreeNode>();
        if (start > end) {
            subTree.Add(null);
            return subTree;
        }
        for (int i = start; i <= end; i++) {
            List<TreeNode> left = dfs(start, i - 1);
            List<TreeNode> right = dfs(i + 1, end);
            for (int j = 0; j < left.Count; j++) {
                for (int k = 0; k < right.Count; k++) {
                    TreeNode tmp = new TreeNode(i);
                    tmp.left = left[j];
                    tmp.right = right[k];
                    subTree.Add(tmp);
                }
            }
        }
        return subTree;
    }
}