【LeetCode】894. All Possible Full Binary Trees 解题报告（Python & C++）

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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

题目大意

给出了个N，代表一棵二叉树有N个节点，求所能构成的树。

解题方法

所有能构成的树，并且返回的不是数目，而是真正的树。所以一定会把所有的节点都求出来。一般就使用了递归。

这个题中，重点是返回一个列表，也就是说每个能够成的树的根节点都要放到这个列表里。而且当左子树、右子树的节点个数固定的时候，也会出现排列组合的情况，所以使用了两重for循环来完成所有的左右子树的组合。

另外的一个技巧就是，左右子树的个数一定是奇数个。

代码如下：

# Definition for a binary tree node.

# class TreeNode(object):

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution(object):

    def allPossibleFBT(self, N):

        """

        :type N: int

        :rtype: List[TreeNode]

        """

        N -= 1

        if N == 0: return [TreeNode(0)]

        res = []

        for l in range(1, N, 2):

            for left in self.allPossibleFBT(l):

                for right in self.allPossibleFBT(N - l):

                    node = TreeNode(0)

                    node.left = left

                    node.right = right

                    res.append(node)

        return res

C++版本的代码如下：

/**

 * Definition for a binary tree node.

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

public:

    vector<TreeNode*> allPossibleFBT(int N) {

        N--;

        vector<TreeNode*> res;

        if (N == 0) {

            res.push_back(new TreeNode(0));

            return res;

        }

        for (int i = 1; i < N; i += 2) {

            for (auto& left : allPossibleFBT(i)) {

                for (auto& right : allPossibleFBT(N - i)) {

                    TreeNode* root = new TreeNode(0);

                    root->left = left;

                    root->right = right;

                    res.push_back(root);

                }

            }

        }

        return res;

    }

};

参考资料：https://leetcode.com/problems/all-possible-full-binary-trees/discuss/163429/Simple-Python-recursive-solution.

日期

2018 年 8 月 26 日 —— 珍爱生命，远离DD！
2018 年 12 月 2 日 —— 又到了周日