# [LeetCode] 894. All Possible Full Binary Trees

linspiration

## Problem

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

Return a list of all possible full binary trees with N nodes. Each element of the answer is the root node of one possible tree.

Each node of each tree in the answer must have node.val = 0.

You may return the final list of trees in any order.

Example 1:

Input: 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]]
Explanation:

Note:

1 <= N <= 20

## Solution

``````class Solution {
public List<TreeNode> allPossibleFBT(int N) {
List<TreeNode> res = new ArrayList<>();
if (N < 1) return res;
if (N == 1) {
return res;
}
if (N%2 == 0) return res;
for (int i = 1; i < N-1; i+=2) {
List<TreeNode> left = allPossibleFBT(i);
List<TreeNode> right = allPossibleFBT(N-1-i);
for (TreeNode l: left) {
for (TreeNode r: right) {
TreeNode root = new TreeNode(0);

root.left = l;
root.right = r;
}
}
}
return res;
}

//If we only need the count
Map<Integer, Integer> map = new HashMap<>();
public int allPossibleCount(int N) {
map.put(1, 1);
if (N == 1) return 1;
if (N < 1 || N%2 == 0) return 0;
int count = 0;
for (int i = 1; i < N-1; i+=2) {
int left = map.containsKey(i) ? map.get(i) : allPossibleCount(i);
int right = map.containsKey(N-1-i) ? map.get(N-1-i) : allPossibleCount(N-1-i);
map.put(i, left);
map.put(N-1-i, right);
count += left * right;
}
return count;
}
}``````