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 };

 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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posted @ 2013-04-22 11:12 ying_vincent 阅读(155) 评论(0) 收藏举报

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