<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:
    void generate(int n, vector<vector<TreeNode*>>&vec){
        for(int i=1;i<n-1;i+=2){
            for(int j=0;j<vec[i].size();j++){
                for(int k=0;k<vec[n-i-1].size();k++){
                    TreeNode* root = new TreeNode(0);
                    root->left = dupli(vec[i][j]);
                    root->right = dupli(vec[n-i-1][k]);
                    vec[n].push_back(root);
                }
            }
        }
    }
    TreeNode* dupli(TreeNode* root){
        if(root){
            TreeNode* root1 = new TreeNode();
            root1->left = dupli(root->left);
            root1->right = dupli(root->right);
            return root1;
        }
        return root;
    }
    vector<TreeNode*> allPossibleFBT(int n) {
        if(n%2 == 0)
            return {};
        vector<vector<TreeNode*>> vec(n+1);
        TreeNode* root = new TreeNode();
        vec[0].push_back(NULL);
        vec[1].push_back(root);
        for(int i=2;i<=n;i++){
            if(i%2 == 0)
                vec[i].push_back(NULL);
            else
                generate(i,vec);
        }
        return vec[n];
    }
};
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