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

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     vector<TreeNode *> dfs(int start, int end) {

13         vector<TreeNode *> subtree;

14         if (start > end) {

15             subtree.push_back(NULL);

16             return subtree;

17         }

18         for (int i = start; i <= end; i++) {

19             vector<TreeNode *> left = dfs(start, i-1);

20             vector<TreeNode *> right = dfs(i+1, end);

21             for (int j = 0; j < left.size(); j++) {

22                 for (int k = 0; k < right.size(); k++) {

23                     TreeNode *tmp = new TreeNode(i);

24                     tmp->left = left[j];

25                     tmp->right = right[k];

26                     subtree.push_back(tmp);

27                 }

28             }

29         }

30         return subtree;

31     }

32     vector<TreeNode *> generateTrees(int n) {

33         // Start typing your C/C++ solution below

34         // DO NOT write int main() function

35         return dfs(1, n);

36     }

37 };

 C#

 1 /**

 2  * Definition for a binary tree node.

 3  * public class TreeNode {

 4  *     public int val;

 5  *     public TreeNode left;

 6  *     public TreeNode right;

 7  *     public TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     public List<TreeNode> GenerateTrees(int n) {

12         return dfs(1, n);

13     }

14     public List<TreeNode> dfs(int start, int end) {

15         List<TreeNode> subTree = new List<TreeNode>();

16         if (start > end) {

17             subTree.Add(null);

18             return subTree;

19         }

20         for (int i = start; i <= end; i++) {

21             List<TreeNode> left = dfs(start, i - 1);

22             List<TreeNode> right = dfs(i + 1, end);

23             for (int j = 0; j < left.Count; j++) {

24                 for (int k = 0; k < right.Count; k++) {

25                     TreeNode tmp = new TreeNode(i);

26                     tmp.left = left[j];

27                     tmp.right = right[k];

28                     subTree.Add(tmp);

29                 }

30             }

31         }

32         return subTree;

33     }

34 }

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