894. All Possible Full Binary Trees**

894. All Possible Full Binary Trees**

https://leetcode.com/problems/all-possible-full-binary-trees/

题目描述

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

C++ 实现 1

基本思路是, 先构建左子树, 再构建右子树, 最后利用根节点连接两棵子树. 需要注意的是, 要生成 full binary tree, 元素个数必须是奇数. 递归到底的情况是, 当输入 N = 1 时. 另外注意到下面代码中, 随着 i 增大, N - 1 - i 可能已经访问过, 为了避免重复访问, 在 C++ 实现 2 引入了额外的空间保存已经得到结果, 可以加快一些速度.

class Solution {
public:
    vector<TreeNode*> allPossibleFBT(int N) {
        if (N % 2 == 0) return {}; 
        if (N == 1) return {new TreeNode(0)};
        vector<TreeNode*> res;
        // i 从 1 开始, 是因为根节点用去了一个节点, 剩下 N - 1 个节点
        for (int i = 1; i <= N - 1; i += 2) {
            if (N - 1 - i >= 1) {
                auto left_candidates = allPossibleFBT(i);
                auto right_candidates = allPossibleFBT(N - 1 - i);
                for (auto &l : left_candidates) {
                    for (auto &r : right_candidates) {
                        TreeNode *root = new TreeNode(0);
                        root->left = l;
                        root->right = r;
                        res.push_back(root);
                    }
                }
            }
        }
        return res;
    }
};

C++ 实现 2

在 C++ 实现 1 代码的基础上, 引入哈希表, 保存已经访问过的结果.

class Solution {
private:
    unordered_map<int, vector<TreeNode*>> record;
public:
    vector<TreeNode*> allPossibleFBT(int N) {
        if (N % 2 == 0) return {}; 
        if (!record.count(N)) {
            if (N == 1) return record[N] = {new TreeNode(0)};
            vector<TreeNode*> res;
            for (int i = 1; i <= N - 1; i += 2) {
                if (N - 1 - i >= 1) {
                    auto left_candidates = allPossibleFBT(i);
                    auto right_candidates = allPossibleFBT(N - 1 - i);
                    for (auto &l : left_candidates) {
                        for (auto &r : right_candidates) {
                            TreeNode *root = new TreeNode(0);
                            root->left = l;
                            root->right = r;
                            res.push_back(root);
                        }
                    }
                }
            }
            record[N] = res;
        }
        return record[N];
    }
};