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生成所有可能的搜索二叉树

时间:2019-02-03 00:54:32      阅读:195      评论:0      收藏:0      [点我收藏+]

标签:cout   \n   ast   new t   bsp   data   int   printf   节点   

product_all_tree(0, 5);

递归返回子树的指针集合。作为 左子树或右子树。

从而构建整颗树。

#include<iostream>
#include<vector>
using namespace std;
//double p[1000]{ 0,0.15,0.10,0.05,0.10,0.20 };
double p[1000]{ 1,2,3,4,5,6,7,8,9 };

#define N 5
class Tree
{
public:
	Tree*parent_;
	Tree*l_c_;
	Tree*r_c_;
	double data_;
	Tree(double a) {
		parent_ = nullptr;
		l_c_ = nullptr;
		r_c_ = nullptr;
		data_ = a;
	}
};
vector<Tree*> product_all_tree(int start,int end)//start end 起始下标,结束下标
{
	vector<Tree*> trees;
	if (start > end)
	{
		trees.push_back(nullptr);
		return trees;
	}
	if (start ==end)
	{
		trees.push_back(new Tree(p[start]));
		return trees;
	}
      //以i作为分界点获取左右两部分的子树集合 for (int i = start; i <= end; ++i) { auto lefts = product_all_tree(start,i-1); auto rights = product_all_tree(i+1, end); for(auto left: lefts) for (auto right : rights) { Tree* root = new Tree(p[i]); root->l_c_ = left; root->r_c_ = right; trees.push_back(root); } } return trees; } void PrintTree(Tree*T, int Layer=1) {/*按竖向树状打印的二叉树*/ int i; if (T == NULL) return; PrintTree(T->r_c_, Layer + 1); for (i = 0; i<Layer; i++) printf(" "); printf("%d\n", static_cast<int>(T->data_)); //按逆中序输出结点,用层深决定结点的左右位置 PrintTree(T->l_c_, Layer + 1); } int main() { auto trees=product_all_tree(0, 5);//一共六个数 for (auto tree : trees) { PrintTree(tree); cout << "-----------------"<< endl ; } }

  结果:二叉树的输出有点丑 横向的,主要关注点,这是个搜索二叉树,

  • 当前根节点的值大于左子树节点的值
  • 当前根节点的值小于右子树节点的值
  • 左右子树同样是二叉搜索树
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请按任意键继续. . .

  

 

cout << "--------------------------" << endl ;

 

生成所有可能的搜索二叉树

标签:cout   \n   ast   new t   bsp   data   int   printf   节点   

原文地址:https://www.cnblogs.com/l2017/p/10349473.html

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