LeetCode: 894. 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]]

Note:

1 <= N <= 20

解题思路 —— 递归求解

将节点按照可能的情况划分给左右子树，递归地求出所有的可能性。

AC 代码

/** * 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 {
    // 拷贝二叉树
    TreeNode* CopyTree(TreeNode* root)
    {
        if(root == nullptr) return nullptr;

        TreeNode* copyRoot = new TreeNode(root->val);
        copyRoot->left = CopyTree(root->left);
        copyRoot->right = CopyTree(root->right);

        return copyRoot;
    }
public:
    vector<TreeNode*> allPossibleFBT(int N) {
        // 偶数节点不可能存在 FBT。
        if( N % 2 == 0)
        {
            return {};
        }

        TreeNode* curRoot = new TreeNode(0);
        vector<TreeNode*> ans;
        --N;
        if(N == 0) return {curRoot};

        for(int i = 1; i < N; i += 2)
        {
            // 分配左右子树的节点数
            vector<TreeNode*> leftTree = allPossibleFBT(i);
            vector<TreeNode*> rightTree = allPossibleFBT(N-i);

            for(int i = 0; i < leftTree.size(); ++i)
            {
                for(int j = 0; j < rightTree.size(); ++j)
                {
                    curRoot->left = leftTree[i];
                    curRoot->right = rightTree[j];
                    ans.push_back(CopyTree(curRoot));
                }
            }
        }

        return ans;
    }
};