Unique Binary Search Trees II
Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.
For example,
Given n = 3, your program should return all 5 unique BST's shown below.
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
Hide Tags Tree Dynamic Programming
SOLUTION 1:
使用递归来做。
1. 先定义递归的参数为左边界、右边界，即1到n.
2. 考虑从left, 到right 这n个数字中选取一个作为根，余下的使用递归来构造左右子树。
3. 当无解时，应该返回一个null树，这样构造树的时候，我们会比较方便，不会出现左边解为空，或是右边解为空的情况。
4. 如果说左子树有n种组合，右子树有m种组合，那最终的组合数就是n*m. 把这所有的组合组装起来即可
1 /**
2 * Definition for binary tree
3 * public class TreeNode {
4 * int val;
5 * TreeNode left;
6 * TreeNode right;
7 * TreeNode(int x) { val = x; left = null; right = null; }
8 * }
9 */
10 public class Solution {
11 public List<TreeNode> generateTrees(int n) {
12 // 0.07
13 return dfs(1, n);
14 }
15
16 public List<TreeNode> dfs(int left, int right) {
17 List<TreeNode> ret = new ArrayList<TreeNode>();
18
19 // The base case;
20 if (left > right) {
21 ret.add(null);
22 return ret;
23 }
24
25 for (int i = left; i <= right; i++) {
26 List<TreeNode> lTree = dfs(left, i - 1);
27 List<TreeNode> rTree = dfs(i + 1, right);
28 for (TreeNode nodeL: lTree) {
29 for (TreeNode nodeR: rTree) {
30 TreeNode root = new TreeNode(i);
31 root.left = nodeL;
32 root.right = nodeR;
33 ret.add(root);
34 }
35 }
36 }
37
38 return ret;
39 }
40 }
CODE:
来源：https://www.cnblogs.com/yuzhangcmu/p/4256291.html