<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:
    TreeNode* clone(TreeNode* root) {
        if(!root) return root;
        TreeNode* node = new TreeNode(root->val);
        node->left = clone(root->left);
        node->right = clone(root->right);
        return node;
    }
    vector<vector<TreeNode*>>dp;
    vector<TreeNode*> allPossibleFBT(int n, int init = true) {
        if(n%2==0) return { };
        if(n==1) return {new TreeNode(0)};
        if(init) dp =vector<vector<TreeNode*>>(n+1);
        if(dp[n].size() > 0) return dp[n];
        vector<TreeNode*> sol;
        for(int i = 1; i < n; i+=2) {
            auto left = allPossibleFBT(i,false);
            auto right = allPossibleFBT(n-1-i,false);
            for(auto l : left) {
                for(auto r: right) {
                    TreeNode* root = new TreeNode(0);
                    root->left = clone(l);
                    root->right = clone(r);
                    sol.push_back(root);
                }
            }
        }
        return dp[n]=sol;
    }
};
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