<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 {
 public:
  vector<TreeNode*> allPossibleFBT(int n) {
    if (n % 2 == 0)
      return {};
    if (n == 1)
      return {new TreeNode(0)};
    if (const auto it = mem.find(n); it != mem.cend())
      return it->second;

    vector<TreeNode*> ans;

    for (int leftCount = 0; leftCount < n; ++leftCount) {
      const int rightCount = n - 1 - leftCount;
      for (TreeNode* left : allPossibleFBT(leftCount))
        for (TreeNode* right : allPossibleFBT(rightCount)) {
          ans.push_back(new TreeNode(0));
          ans.back()->left = left;
          ans.back()->right = right;
        }
    }

    return mem[n] = ans;
  }

 private:
  unordered_map<int, vector<TreeNode*>> mem;
};
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