<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*> helper(int n,vector<vector<TreeNode*>>&dp){
      if(dp[n].size()){return dp[n];}
      if(n == 0 || n==2){return {};}
      if(n==1){ TreeNode* temp = new TreeNode(0); return {temp};}   
      vector<TreeNode*>ans;    
      for(int j=0;j<n;j++){
         vector<TreeNode*>lefts = helper(j,dp);
         vector<TreeNode*>rights = helper(n-j-1,dp);
         cout<<n<<" "<<lefts.size()<<" "<<rights.size()<<endl;
         for(int i=0;i<lefts.size();i++){
            for(int k=0;k<rights.size();k++){
                TreeNode* newhead = new TreeNode(0);
                newhead->left = lefts[i];
                newhead->right = rights[k];
                ans.push_back(newhead);
            }
         }
      }
      dp[n] = ans;
      return ans;


    }
    vector<TreeNode*> allPossibleFBT(int n) {
        vector<TreeNode*>ans;
        vector<vector<TreeNode*>>dp(n+1,vector<TreeNode*>(0));
        return helper(n,dp);
        
        
    }
};
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