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.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func allPossibleFBT(n int) []*TreeNode {
	f := make([][]*TreeNode, n+1)
	var dfs func(int) []*TreeNode
	dfs = func(n int) []*TreeNode {
		if len(f[n]) > 0 {
			return f[n]
		}
		if n == 1 {
			return []*TreeNode{&TreeNode{Val: 0}}
		}
		ans := []*TreeNode{}
		for i := 0; i < n-1; i++ {
			j := n - 1 - i
			for _, left := range dfs(i) {
				for _, right := range dfs(j) {
					ans = append(ans, &TreeNode{0, left, right})
				}
			}
		}
		f[n] = ans
		return ans
	}
	return dfs(n)
}