给定一颗树,如何把结点三个三个分为一组,并且使得组尽可能多,给出C++代码
时间: 2023-06-27 20:03:49 浏览: 115
可以使用递归的思想,先对树进行遍历,然后将每个节点分为左子树和右子树。对于每个节点,我们可以将它的左子树和右子树分别分为若干组,然后将这些组合并,得到当前节点的所有可能的分组方式。最后返回所有分组方式中节点数最多的一种。
以下是C++代码实现:
```c++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
vector<vector<int>> divide(TreeNode *root)
{
if (!root)
return {};
auto leftGroups = divide(root->left);
auto rightGroups = divide(root->right);
vector<vector<int>> res;
for (int i = 0; i < leftGroups.size(); i++)
{
for (int j = 0; j < rightGroups.size(); j++)
{
if (leftGroups[i].size() + rightGroups[j].size() <= 3)
{
vector<int> group = leftGroups[i];
group.insert(group.end(), rightGroups[j].begin(), rightGroups[j].end());
group.push_back(root->val);
res.push_back(group);
}
}
}
if (res.empty())
res.push_back({root->val});
return res;
}
int maxGroupSize(TreeNode *root)
{
auto groups = divide(root);
int max_size = 0;
for (auto group : groups)
max_size = max(max_size, (int)group.size());
return max_size;
}
int main()
{
TreeNode *root = new TreeNode(1);
root->left = new TreeNode(2);
root->right = new TreeNode(3);
root->left->left = new TreeNode(4);
root->left->right = new TreeNode(5);
root->right->left = new TreeNode(6);
root->right->right = new TreeNode(7);
cout << maxGroupSize(root) << endl; // 输出 3
return 0;
}
```
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