Unique Binary Search Trees II
<p>Given an integer <code>n</code>, return <em>all the structurally unique <strong>BST&#39;</strong>s (binary search trees), which has exactly </em><code>n</code><em> nodes of unique values from</em> <code>1</code> <em>to</em> <code>n</code>. Return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0095.Unique%20Binary%20Search%20Trees%20II/images/uniquebstn3.jpg" style="width: 600px; height: 148px;" />
<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [[1]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 8</code></li>
</ul>
Tree; Binary Search Tree; Dynamic Programming; Backtracking; Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        return dfs(1, n);
    }

    private List<TreeNode> dfs(int i, int j) {
        List<TreeNode> ans = new ArrayList<>();
        if (i > j) {
            ans.add(null);
            return ans;
        }
        for (int v = i; v <= j; ++v) {
            var left = dfs(i, v - 1);
            var right = dfs(v + 1, j);
            for (var l : left) {
                for (var r : right) {
                    ans.add(new TreeNode(v, l, r));
                }
            }
        }
        return ans;
    }
}