# All Possible Full Binary Trees Problem

## Description

LeetCode Problem 894.

Given an integer n, return a list of all possible full binary trees with n nodes. Each node of each tree in the answer must have Node.val == 0.

Each element of the answer is the root node of one possible tree. You may return the final list of trees in any order.

A full binary tree is a binary tree where each node has exactly 0 or 2 children.

Example 1:

``````1
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Input: n = 7
Output: [[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]]
``````

Example 2:

``````1
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Input: n = 3
Output: [[0,0,0]]
``````

Constraints:

• 1 <= n <= 20

## Sample C++ Code

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/**
* 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*> allPossibleFBT(int N) {
vector<TreeNode*> ans;
TreeNode* root = new TreeNode(0);
if (N == 1) {
ans.push_back(root);
return ans;
} else if (N % 2 == 1) {
for (int x = 0; x < N; x ++) {
int y = N - x - 1;
vector<TreeNode*> left_ans = allPossibleFBT(x);
vector<TreeNode*> right_ans = allPossibleFBT(y);
for (vector<TreeNode*>::iterator it1 = left_ans.begin();
it1 != left_ans.end(); it1 ++) {
for (vector<TreeNode*>::iterator it2 = right_ans.begin();
it2 != right_ans.end(); it2 ++) {
TreeNode* top = new TreeNode(0);
top->left = *it1;
top->right = *it2;
ans.push_back(top);
}
}
}
}
return ans;
}
};
``````