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
分析：对n，合法的BST有C(n,2*n)/(n+1))种.
依照后序建树，[1，..i⑴] i [i+1,..n]先分别建好左右合法的BST，然后接到root上。
设左子树形态有m种，右子树有k种，那末要尝试m*k种拼接方式。
public List<TreeNode> generateTrees(int l, int r) {
List<TreeNode> list = new ArrayList<TreeNode>();
if (l > r) {
list.add(null);
return list;
}
for (int i = l; i <= r; i++) {
List<TreeNode> lefts = generateTrees(l, i - 1);
List<TreeNode> rights = generateTrees(i + 1, r);
for (TreeNode left : lefts) {
for (TreeNode right : rights) {
TreeNode root = new TreeNode(i);
root.left = left;
root.right = right;
list.add(root);
}
}
}
return list;
}
public List<TreeNode> generateTrees(int n) {
return generateTrees(1, n);
}