基本二叉树程序C++

Question

所以我一直在学习所有关于二叉树的知识，并决定编写一个简单的程序来向我自己证明我可以将我的知识应用到工作代码中。我试图用这个代码做的事情是在一棵二叉树中添加4个数字，并按照从最小到最大的顺序输出数字。虽然，我的代码遇到了问题。当我运行代码时，Visual Studio在第29行和第59行将其分开。我相信问题与递归函数addLeaf有关，但也许是其他内容。任何建议，解决方案，或输入将不胜感激。！

#include "stdafx.h" 
#include <iostream> 
#include <cstdlib> 
#include <fstream> 
using namespace std; 

struct node 
{ 
    int data; 
    node* left; 
    node* right; 
}; 

node* root = NULL; 

node* createLeaf(int data) 
{ 
    node* n = new node; 
    n->data = data; 
    n->left = NULL; 
    n->right = NULL; 

    return n; 
} 
void addLeaf(int data) 
{ 
    node* curr = root; 


    //If tree is empty, create first node 
    if(root == NULL) 
    { 
     root = createLeaf(data); 

    } 

    //Left(Less than) 
    else if(data < curr->data) 
    { 
     //Check for curr->left 
     if(curr->left != NULL) 
     { 
      addLeaf(data); 
     } 
     else //Adds node to left if null 
     { 
      curr->left = createLeaf(data); 
     } 
    } 
    //Right(greater than) 
    else if(data > curr->data) 
    { 
     //Check for curr->right 
     if(curr->right != NULL) 
     { 
      addLeaf(data); 
     } 
     else //Adds node if right is Null 
     { 
      curr->right = createLeaf(data); 
     } 
    } 
    else 
    { 
     cout << "The data " << data << " has already been received\n"; 
    } 


} 

void printTree(node* Ptr) 
{ 


    if(root != NULL) 
    { 
     if(Ptr->left != NULL) 
     { 
      printTree(Ptr->left); 
     } 
     cout << Ptr->data << " "; 
     if(Ptr->right != NULL) 
     { 
      printTree(Ptr->right); 
     } 
     cout << Ptr->data << " "; 
    } 
    else 
    { 
     cout << "The Tree is empty\n"; 
    } 


} 

int main() 
{ 
    int data[4] = {1, 7, 5, 4}; 
    node* Ptr = root; 


    for(int i = 0; i < 4; i++) 
    { 
     addLeaf(data[i]); 
    } 

    printTree(Ptr); 



    system("PAUSE"); 
    return 0; 
} 

Answer 1

addleaf函数将无限运行。您只需不加任何检查地添加到根目录。 您将Ptr指定为root，但后来使用new，将其分配给内存中某个根并未指向的新地址。 您必须通过Ptr才能参照addLeaf，否则将对其副本进行更改，该副本将在addLeaf终止时销毁。 printTree打印当前节点值的两倍（复制粘贴错误？）

下面是完整的代码：

#include "stdafx.h" 
#include <iostream> 
#include <cstdlib> 
#include <fstream> 
using namespace std; 

struct node 
{ 
    int data; 
    node* left; 
    node* right; 
}; 

node* root = NULL; 

node* createLeaf(int data) 
{ 
    node* n = new node; 
    n->data = data; 
    n->left = NULL; 
    n->right = NULL; 

    return n; 
} 
void addLeaf(node* &curr, int data) 
{ 
    //If tree is empty, create first node 
    if(curr == NULL) 
    { 
     curr = createLeaf(data); 
    } 

    //Left(Less than) 
    else if(data < curr->data) 
    { 
     addLeaf (curr->left, data); 
    } 
    //Right(greater than) 
    else if(data > curr->data) 
    { 
     addLeaf(curr->right, data); 
    } 
    else 
    { 
     cout << "The data " << data << " has already been received\n"; 
    } 
} 

void printTree(node* Ptr) 
{ 


    if(root != NULL) 
    { 
     if(Ptr->left != NULL) 
     { 
     printTree(Ptr->left); 
     } 
     cout << Ptr->data << " "; 
     if(Ptr->right != NULL) 
     { 
     printTree(Ptr->right); 
     } 
    } 
    else 
    { 
     cout << "The Tree is empty\n"; 
    } 


} 

int main() 
{ 
    int data[4] = {1, 7, 5, 4}; 

    for(int i = 0; i < 4; i++) 
    { 
     addLeaf(root, data[i]); 
    } 

    printTree(root); 

    system("PAUSE"); 
    return 0; 
} 

Answer 2

一个问题，我可以发现：

void addLeaf(int data) 
{ 
    node* curr = root; 
..... 
     //Check for curr->left 
     if(curr->left != NULL) 
     { 
      addLeaf(data); 
     } 

你所谓的递归什么也没做。它只会继续调用addLeaf函数，并且函数继续检查root的左边是否为空，并再次调用addLeaf。

重构所有的代码。不要使用任何全局变量。确保您传递了正确的参数（例如，您应该将下一级节点传递给addLeaf）