我创建了一个包含类Node的程序,用于表示任何类型(模板)的二叉树。
在我的Node.h类中,我有两个构造函数,但是不确定我是否正确实现了这两个构造函数。在构造函数中初始化值使我感到困惑。在我的main.cpp文件中,我有一个setUpTree函数。我的程序现在执行,但是不打印设置的树。
我已经尝试了几个小时试图解决这个问题,但是没有止境。我对C ++,指针,构造函数等还没有真正的经验。
如果有人可以帮助我修复代码以便setUpTree函数以及printTree方法起作用,我将不胜感激。
谢谢
Node.h类:
#ifndef NODE_H
#define NODE_H
#include <iostream>
#include <string>
using namespace std;
//an object of type node holds 3 things
// - an item (of type t)
// - a left subtree
// - a right subtree
template<typename T>
class Node {
public:
Node(T item); //constructor to create a leaf node
Node(T item, Node *lft, Node *rht); //constructor which creates an internal node
~Node(); //Destructor
//public data member functions:
bool searchTree(T key);
void printTree();
private:
//private data member functions:
Node* left;
Node* right;
T item;
};
//constructor
template<typename T>
Node<T>::Node(T i, Node<T> *lft, Node<T> *rht) {
item = i;
left = NULL;
right = NULL;
}
//constructor
template <typename T>
Node<T>::Node(T i) { //should i be a parameter here?
item = i; //is this right for this constructor?
}
//destructor
template <typename T>
Node<T>::~Node() {
delete left;
delete right;
//delete;
}
//print tree method
template <typename T>
void Node<T>::printTree() {
if (left != NULL) {
left->printTree();
cout << item << endl;//alphabetical order
}
if (right != NULL) {
right->printTree();
//cout << item << endl; //post order
}
}
//search Tree method
template <typename T>
bool Node<T>::searchTree(T key) {
bool found = false;
if (item == key) {
return true;
}
if (left != NULL) {
found = left->searchTree(key);
if (found) return true;
}
if (right != NULL) {
return right->searchTree(key);
}
return false; //if left and right are both null & key is not the search item, then not found == not in the tree.
}
#endif
Main.cpp类别:
#include "Node.h"
#include <iostream>
using namespace std;
//set up tree method
Node<string> *setUpTree() {
Node<string> *s_tree =
new Node<string>("Sunday",
new Node<string>("monday",
new Node<string>("Friday"),
new Node<string>("Saturday")),
new Node<string>("Tuesday",
new Node<string>("Thursday"),
new Node<string>("Wednesday")));
return s_tree;
}
int main() {
Node<string> *s_tree;
s_tree = setUpTree(); //call setUpTree method on s_tree
cout << "Part 2 :Printing tree values: " << endl;
s_tree->printTree(); //call print tree method
cout << endl;
//search for range of tree values
//searchTree(s_tree, "Sunday");
//searchTree(s_tree, "Monday");
return 0;
}
最佳答案
我不知道这是否是唯一的问题,但是...如果构造叶子,则必须将left
和right
指针设置为NULL
template <typename T>
Node<T>::Node(T i) : left(NULL), right(NULL), item(i)
{ }
否则,在调用析构函数时
template <typename T>
Node<T>::~Node() {
delete left;
delete right;
//delete;
}
delete
在未定义的值上被调用;两次。这是崩溃的完美秘诀。
在使用
left
或right
的每个点中,还存在其他问题,如NULL
或printTree()
那样检查指针是否为searchTree()
:该值是不确定的,因此可以是非NULL
,请通过测试并且printTree()
在具有未定义值的指针上调用-编辑-
建议的构造函数。
template <typename T>
Node<T>::Node (T i, Node<T> * lft, Node<T> * rht)
: left(lft), right(right), item(i)
{ }
template <typename T>
Node<T>::Node (T i)
: left(NULL), right(NULL), item(i)
{ }
-编辑2-
现在至少打印一些值;星期一星期天星期二星期二。不确定其余
看看你的
printTree()
方法模板
无效Node :: printTree(){
if(left!= NULL){
left-> printTree();
cout < }
if (right != NULL) {
right->printTree();
//cout << item << endl; //post order
}
}
仅当
item
不是left
时才打印值(NULL
)。因此,它不会显示叶子的值。建议:即使
printTree()
是item
,也要修改left
以打印NULL
。举个例子
template <typename T>
void Node<T>::printTree() {
if (left != NULL) {
left->printTree();
}
cout << item << endl;
if (right != NULL) {
right->printTree();
}
}
关于c++ - 设置二叉树,打印并搜索的程序-Node Class C++,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/41147667/