<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*> nodes;
  vector<TreeNode*> res;
  
  TreeNode* clone(TreeNode* z) {
    if (z == NULL) return NULL;
    TreeNode* tmp = new TreeNode();
    tmp->left = clone(z->left);
    tmp->right = clone(z->right);
    return tmp;
  }
  
  void gen(int n, int i) {
    if (i == (int) nodes.size()) {
      if (i == n) res.push_back(clone(nodes[0]));
      return;
    }
    
    // Make nodes[i] a 2 children node.
    if (nodes.size() + 2 <= n) {
      nodes[i]->left = new TreeNode(nodes.size());
      nodes.push_back(nodes[i]->left);
      nodes[i]->right = new TreeNode(nodes.size());
      nodes.push_back(nodes[i]->right);
      gen(n, i+1);
      nodes[i]->left = NULL;
      nodes.pop_back();
      nodes[i]->right = NULL;
      nodes.pop_back();
    }
    
    // Make nodes[i] a 0 children node.
    gen(n, i+1);
  }
  
  vector<TreeNode*> allPossibleFBT(int n) {
    res.clear();
    nodes.push_back(new TreeNode());
    gen(n, 0);
    return res;
  }
};
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