<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>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> allPossibleFBT(int n) {
        static map<int, vector<TreeNode*>> tree_maps;
        if (n % 2 == 0) {
            return vector<TreeNode*>();
        }
        if (tree_maps.find(n) != tree_maps.end()) {
            return tree_maps.at(n);
        }
        auto& result = tree_maps[n];
        if (n == 1) {
            result.emplace_back(new TreeNode());
        } else {
            for (int left = 1, right = n - 1 - left; left <= right; left += 2, right -= 2) {
                auto left_trees = allPossibleFBT(left);
                auto right_trees = allPossibleFBT(right);
                for (auto left_root : left_trees) {
                    for (auto right_root : right_trees) {
                        result.emplace_back(new TreeNode(0, left_root, right_root));
                        if (left != right) {
                            result.emplace_back(new TreeNode(0, right_root, left_root));
                        }
                    }
                }
            }
        }
        return result;
    }
};
31826