如何修复c++中的分段错误11

Question

我目前正在用c++编写一个二进制树项目，它给了我一些难以理解的错误。有问题的错误是一个分段错误，我不知道如何修复。以下是其中一个文件的代码，所有函数都是在其中编写的。特别是给我带来问题的函数是最后一个——print_pre_order_private函数。

#include <iostream>
#include <cstdlib>
#include "BT.h"
using namespace std;
BT::BT()
{
root = NULL;  //makes sure pointer isn't pointing at anything
}
BT::node* BT::create_leaf(int key)
{
node* n = new node;
n->key = key;
n->left = NULL;
n->right = NULL;
return n;
}
void BT::add_leaf(int key)
{
add_leaf_private(key, root);
} 
void BT::add_leaf_private(int key, node* Ptr)
{
if(root == NULL)
{
root = create_leaf(key);
}
else if(key < Ptr->key)
{
if(Ptr->left != NULL)
{
add_leaf_private(key, Ptr->left);
}
else
{
Ptr->left = create_leaf(key);
}
}
else if(key > Ptr->key)
{
if(Ptr->right != NULL)
{
add_leaf_private(key, Ptr->right);
}
else
{
Ptr->right = create_leaf(key);
}
}
else
{
cout << "The key " << key << " has already been added to the    treen";
}
}
void BT::print_in_order()
{
print_in_order_private(root);
}
void BT::print_in_order_private(node* Ptr)
{
if(root != NULL)
{
if(Ptr->left != NULL)
{
print_in_order_private(Ptr->left);
} 
cout << Ptr->key << " ";
if(Ptr->right != NULL)
{
print_in_order_private(Ptr->right);
}
} 
else
{
cout << "The tree is emptyn";
}
}
BT::node* BT::return_node(int key)
{
return return_node_private(key, root);
}
BT::node* BT::return_node_private(int key, node* Ptr)
{
if(Ptr != NULL)
{
if(Ptr->key == key)
{
return Ptr;
}
else
{
if(key < Ptr->key)
{
return return_node_private(key, Ptr->left);
}
else
{
return return_node_private(key, Ptr->right);
}
}
}
else
{
return NULL;
}
}
int BT::return_root_key()
{
if(root != NULL)
{
return root->key;
}
else
{
return -1;
}
}
void BT::print_children(int key)
{
node* Ptr = return_node(key);
if(Ptr != NULL)
{
cout << "Parent Node = " << Ptr->key << endl;
Ptr->left == NULL ?
cout << "Left child = NULLn":
cout << "Left child = " << Ptr->left->key << endl;
Ptr->right == NULL ?
cout << "Right child = NULLn":
cout << "Right child = " << Ptr->right->key << endl;
}
else
{
cout << "Key " << key << " is not in the treen";
}
}
int BT::find_smallest()
{
return find_smallest_private(root);
}
int BT::find_smallest_private(node* Ptr)
{
if(root == NULL)
{
cout << "The tree is emptyn";
return -1;
}
else
{
if(Ptr->left != NULL)
{
return find_smallest_private(Ptr->left);
}
else
{
return Ptr->key;
}
}
}

void BT::remove_node(int key)
{
remove_node_private(key, root);
}  
void BT::remove_node_private(int key, node* parent)
{
if(root != NULL)
{
if(root->key == key)
{
remove_root_match();
}
else
{
if(key < parent->key && parent->left != NULL)
{
parent->left->key == key ?
remove_match(parent, parent->left, true) :
remove_node_private(key, parent->left);
}   
else if(key < parent->key && parent->right != NULL)
{
parent->right->key == key ?
remove_match(parent, parent->right, true) :
remove_node_private(key, parent->right);
}
else
{
cout << "The key " << key << " was not found in the treen";
}
}
}
else
{
cout << "The tree is emptyn";
}
}
void BT::remove_root_match()
{
if(root != NULL)
{
node* delPtr = root;
int root_key = root->key;
int smallest_right_subtree;
//case 0 - 0 children
if(root->left == NULL && root->right == NULL)
{
root = NULL;
delete delPtr;
}
//case 1 - 1 child
else if(root->left == NULL && root->right != NULL)
{
root = root->right;
delPtr->right = NULL;
delete delPtr;
cout << "The root node with key " << root_key << " was deleted. " 
<< " The new root contains key " << root->key << endl;
}
else if(root->right == NULL && root->left != NULL)
{
root = root->left;
delPtr->left = NULL;
delete delPtr;
cout << "The root node with key " << root_key << " was deleted. " 
<< " The new root contains key " << root->key << endl;
}
//case 2 - 2 children
else
{
smallest_right_subtree = find_smallest_private(root->right);
remove_node_private(smallest_right_subtree, root);
root->key = smallest_right_subtree;
cout << "The rout key containing key " << root_key
<< " was overwritten with key " << root->key << endl;
}
}
else
{
cout << "Cannot remove root. Tree is emptyn";
}
}
void BT::remove_match(node* parent, node* match, bool left)
{
if(root != NULL)
{
node* delPtr;
int match_key = match->key;
int smallest_right_subtree;
//case 0 - 0 children
if(match->left == NULL && match->right == NULL)
{
delPtr = match;
left == true ? parent->left = NULL : parent->right = NULL;
delete delPtr;
cout << "The node containing key " << match_key << " was removedn";
}
//case 1 - 1 child
else if(match->left == NULL && match->right != NULL)
{
left == true ? parent->left = match->left : parent->right = match->right;
match->left = NULL;
delPtr = match;
delete delPtr;
cout << "The node containing key " << match_key << " was removedn";
}
//case 2 - 2 children
else
{
smallest_right_subtree = find_smallest_private(match->right);
remove_node_private(smallest_right_subtree, match);
match->key = smallest_right_subtree;
}
}
else 
{
cout << "Cannot remove match. The tree is empty";
}
}
void print_pre_order()
{
print_pre_order_private(root);
}
void print_pre_order_private(node* Ptr)
{
if(root != NULL)
{
cout << Ptr->key << " ";
print_pre_order_private(Ptr->left);
print_pre_order_private(Ptr->right);
}
}

我们非常感谢为解决这一问题所提供的任何帮助。提前感谢。：)

Answer 1

In-print_pre_order_private(node*Ptr)函数。这不应该是"if(Ptr！=NULL)"而不是"if(root！=NULL"吗？。我想是撞车了。

由于您已将"print_pre_order_private()"函数设计为递归函数，因此您应该检查"Ptr"的有效性，而不是始终检查"root"的合法性，因为根据您的输入，子项的左/右/两者在树的最底部都可能为NULL。