All Possible Full Binary Trees
<p>Given an integer <code>n</code>, return <em>a list of all possible <strong>full binary trees</strong> with</em> <code>n</code> <em>nodes</em>. Each node of each tree in the answer must have <code>Node.val == 0</code>.</p>

<p>Each element of the answer is the root node of one possible tree. You may return the final list of trees in <strong>any order</strong>.</p>

<p>A <strong>full binary tree</strong> is a binary tree where each node has exactly <code>0</code> or <code>2</code> children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0894.All%20Possible%20Full%20Binary%20Trees/images/fivetrees.png" style="width: 700px; height: 400px;" />
<pre>
<strong>Input:</strong> n = 7
<strong>Output:</strong> [[0,0,0,null,null,0,0,null,null,0,0],[0,0,0,null,null,0,0,0,0],[0,0,0,0,0,0,0],[0,0,0,0,0,null,null,null,null,0,0],[0,0,0,0,0,null,null,0,0]]
</pre>

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

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> [[0,0,0]]
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 20</code></li>
</ul>
Tree; Recursion; Memoization; Dynamic Programming; Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private List<TreeNode>[] f;

    public IList<TreeNode> AllPossibleFBT(int n) {
        f = new List<TreeNode>[n + 1];
        return Dfs(n);
    }

    private IList<TreeNode> Dfs(int n) {
        if (f[n] != null) {
            return f[n];
        }
        
        if (n == 1) {
            return new List<TreeNode> { new TreeNode() };
        }
        
        List<TreeNode> ans = new List<TreeNode>();
        for (int i = 0; i < n - 1; ++i) {
            int j = n - 1 - i;
            foreach (var left in Dfs(i)) {
                foreach (var right in Dfs(j)) {
                    ans.Add(new TreeNode(0, left, right));
                }
            }
        }
        f[n] = ans;
        return ans;
    }
}