894. All Possible Full Binary Trees

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:

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:

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

Constraints:

• 1 <= n <= 20

题目：给定正整数n。求包含n个节点的所有完全二叉树的结构。每个节点值为0. 完全二叉树是一个二叉树，其所有节点包含0个或2个子节点。

思路：与96. Unique Binary Search Trees类似，采用动态规划方法，将n-1个节点分为左右子节点。采用递归方法分别求出从1到n-2个节点的完全二叉树结构，将其置于根节点的左右子树位置。为减少计算，使用hashmap来保存子树结构。小编使用一个vector> subTree结构来代替hashmap，subTree[i]即为包含i个节点的所有完全二叉树结构。为保证树结构为完全二叉树，i为奇数。代码：

/**
 * 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 {
private:
    vector> subTree;
public:
    vector allPossibleFBT(int n) {
        if(subTree.size() > n && subTree[n].size() > 0)
            return subTree[n];
        while(subTree.size() <= n) subTree.push_back({});
        if(n == 1){
            subTree[1].push_back(new TreeNode(0));
        } else {            
            for(int i = 1; i < n-1; i+=2){
                vector left = allPossibleFBT(i);
                vector right = allPossibleFBT(n-i-1);
                for(auto l : left){
                    for(auto r : right){
                        TreeNode* root = new TreeNode(0);
                        root->left = l;
                        root->right = r;
                        subTree[n].push_back(root);
                    }
                }
            }
        }
        return subTree[n];
    }
};