<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:
    // Memoization to store results for subproblems
    vector<vector<TreeNode*>> memo;
    
    // Function to generate all full binary trees with n nodes
    vector<TreeNode*> allPossibleFBT(int n) {
        // If the result for n nodes is already computed, return it
        if (!memo[n].empty()) return memo[n];
        
        // Base case: A full binary tree with 1 node is just a single node
        if (n == 1) return {new TreeNode(0)};
        
        // List to store all possible trees with n nodes
        vector<TreeNode*> result;
        
        // Iterate over all possible left and right subtree sizes
        // Since n is odd, we iterate from 1, 3, 5, ..., n-2
        for (int leftNodes = 1; leftNodes < n; leftNodes += 2) {
            int rightNodes = n - 1 - leftNodes;
            vector<TreeNode*> leftSubtrees = allPossibleFBT(leftNodes);
            vector<TreeNode*> rightSubtrees = allPossibleFBT(rightNodes);
            
            // Combine each left and right subtree with a new root node
            for (TreeNode* left : leftSubtrees) {
                for (TreeNode* right : rightSubtrees) {
                    TreeNode* root = new TreeNode(0, left, right);
                    result.push_back(root);
                }
            }
        }
        
        // Store the result in the memoization table
        memo[n] = result;
        return result;
    }
    
    Solution() {
        // Initialize the memoization table with size 21 (1 <= n <= 20)
        memo.resize(21);
    }
};
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