生成所有可能的搜索二叉树
Posted l2017
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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 ", 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 ;
以上是关于生成所有可能的搜索二叉树的主要内容，如果未能解决你的问题，请参考以下文章
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