題目：894. All Possible Full Binary Trees(https://leetcode.com/problems/all-possible-full-binary-trees/description/).

這個問題是典型的遞迴問題，首先讀題之後可知有性質“完全二叉樹結點個數必定為奇數”。所以問題變簡單了點，如果輸入N為偶數，直接輸出空List即可。

對於一個有n個結點的完全二叉樹，其可以分解為有i個結點（i為奇數）的左子樹和n-i-1個結點的右子樹，所以只需遍歷i的取值即可！

class Solution {
    public List<TreeNode> allPossibleFBT(int N) {
        List<TreeNode> res = new ArrayList<>();
        if (N == 1) {
            res.add(new TreeNode(0));
            return res;
        }
        if (N % 2 == 0) {
            return res;
        }
        for (int i = 1; i < N; i = i + 2) {
            List<TreeNode> lns = allPossibleFBT(i);
            List<TreeNode> rns = allPossibleFBT(N - i - 1);
            for (TreeNode ln : lns) {
                for (TreeNode rn : rns) {
                    TreeNode head = new TreeNode(0);
                    head.left = ln;
                    head.right = rn;
                    res.add(head);
                }
            }
        }
        return res;
    }
}
 

感謝：https://leetcode.com/problems/all-possible-full-binary-trees/discuss/164134/C%2B%2B-solution-easy-to-understand-and-use-chinese%3A)