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.

Have you been asked this question in an interview?

自认为tree 和recursion 掌握的很好了，做这道题目竟然被卡住了。由于这道题目是需要存储 every unique tree，同时 由于是reference 类型，不能直接copy，每次都需要重新new。这是需要注意的地方。同时，recursion 本身是指子问题和原问题是相同解法，因此，这道题目最后创建了 n多 arraylist。每次的root 都要重新new。这两点是解题的关键。

另外要注意一个小细节，arraylist 是可以添加null 值的, 不知道这一点，写起来也会有一点麻烦。

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; left = null; right = null; }
* }
*/
public class Solution {
public List<TreeNode> generateTrees(int n) {
return constructTree(1, n);
}

List<TreeNode> constructTree(int start, int end) {
List<TreeNode> result = new ArrayList<TreeNode> ();
if (start > end) {
result.add(null);
return result;
}

for (int i = start; i <= end; i++) {
List<TreeNode> left = constructTree(start, i - 1);
List<TreeNode> right = constructTree(i + 1, end);
for (TreeNode l: left) {
for (TreeNode r: right) {
TreeNode root = new TreeNode(i);
root.left = l;
root.right = r;
result.add(root);
}
}
}
return result;
}

}