我试图创建一个非递归的insert()函数。我在书中唯一的例子是递归的,我试图转换它。这样你就可以了解我想要完成的事情,以及为什么我会包含这些说明。
编写一个类来实现一个能够存储数字的简单二进制搜索树。该类应具有成员函数:
void insert(double x)
bool search(double x)
void inorder(vector <double> & v)
insert函数不应该通过调用递归函数直接或间接使用递归。
还有更多,但我认为这给出了我所询问的背后的想法。截至目前,该函数只是继续重新创建根节点。这就是我所拥有的。
编辑:为了清晰起见,添加完整代码。
#include "stdafx.h"
#include <iostream>
#include <vector>
class BinaryTree {
private:
struct TreeNode {
double value;
TreeNode *left;
TreeNode *right;
TreeNode(double value1,
TreeNode *left1 = nullptr,
TreeNode *right1 = nullptr) {
value = value1;
left = left1;
right = right1;
}
};
TreeNode *root; //pointer to the root of the tree
bool search(double x, TreeNode *t) {
while (t) {
std::cout << "running through t." << std::endl;
if (t->value == x) {
return true;
}
else if (x < t->value) {
std::cout << "wasn't found, moving left." << std::endl;
search(x, t->left);
}
else {
std::cout << "wasn't found, moving right." << std::endl;
search(x, t->right);
}
}
std::cout << "wasn't found." << std::endl;
return false;
}
public:
std::vector<TreeNode> v;
BinaryTree() {
root = nullptr;
}
void insert(double x) {
TreeNode *tree = root;
if (!tree) {
std::cout << "Creating tree." << x << std::endl;
root = new TreeNode(x);
return;
}
while (tree) {
std::cout << "Adding next value." << std::endl;
if (tree->value == x) return;
if (x < tree->value) {
tree = tree->left;
tree->value = x;
}
else {
tree = tree->right;
tree->value = x;
}
}
}
bool search(double x) {
return search(x, root);
}
/*void inOrder(TreeNode *v) const {
while (root != nullptr) {
inOrder(root->left);
v.push_back(root->value);
inOrder(root->right);
v.push_back(root->value);
}
}*/
};
int main() {
BinaryTree t;
std::cout << "Inserting the numbers 5, 8, 3, 12, and 9." << std::endl;
t.insert(5);
t.insert(8);
t.insert(3);
t.insert(12);
t.insert(9);
std::cout << "Looking for 12 in tree." << std::endl;
if (t.search(12)) {
std::cout << "12 was found." << std::endl;
}
std::cout << "Here are the numbers in order." << std::endl;
return 0;
}
答案 0 :(得分:0)
在insert()
方法中,您可以将传递的每个节点的值设置为新值。这可能不是你想要的。
如果要插入它,需要弄清楚它的位置,创建一个具有适当值的新节点,然后将新节点插入树中,小心处理以前的节点。