<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://s3-lc-upload.s3.amazonaws.com/uploads/2018/08/22/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>

class Solution:
    
    def allPossibleFBT(self, n: int) -> List[Optional[TreeNode]]:
        
        if n == 1:
            return [TreeNode()]
        
        result = []
        
        for i in range(1, n - 1, 2):  
            
            left_subtrees = self.allPossibleFBT(i)  
            right_subtrees = self.allPossibleFBT(n - 1 - i)  
            
            for left in left_subtrees:
                for right in right_subtrees:
                    root = TreeNode()  
                    root.left = left 
                    root.right = right  
                    result.append(root)  
        
        return result  

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