C++ delete node and its child nodes [duplicate] - c++

Question

This question already has answers here:
Binary Search Tree Destructor
(6 answers)
Closed 2 years ago.
Please help me. I am stuck at this.
What am I trying to do: Binary search tree.
I am a C# developer and I learn C++ for about 2 weeks, therefore don't be so harsh with me and that's why pointers are still difficult for me.
I have a struct Node
struct Node
{
int Value;
Node* _LeftNode;
Node* _RightNode;
Node(int value)
: Value(value), _LeftNode(NULL), _RightNode(NULL)
{
}
};
and a Delete() function in BinarySearchTree.cpp
void BinarySearchТрее::Delete(Node* node)
{
if (node)
{
Delete(node->_LeftNode);
Delete(node->_RightNode);
delete(node);
node = NULL;
}
}
I want to delete the node and all of its child nodes.
When I first step in the recursion... For example:
I have two child nodes with values 10 and 19.
With recursion, I delete the nodes and set the pointers to NULL.
And here is the problem:
When I came out from the recursion the nodes are not NULL, but something strange.
And this is my problem. Why when I am in the recursion and I NULL the pointer everything is fine, but when I come out the pointer is something else.

Answer 1

As I talked in the comments, I think the thing is that how we can reset the pointer of the parent's(left or right child) of the initially passed node. (recursively deleting a node and its all children looks good.)
And I don't think it is possible in your current design. As Node does not contain a pointer to its parent, so there is no way to know who's the parent. node = NULL sets just the argument(local variable)'s value so it is pointless.

Answer 2

The C++ way would be to use std::unique_ptr.
struct Node
{
int Value;
std::unique_ptr<Node> LeftNode;
std::unique_ptr<Node> RightNode;
Node(int value)
: Value(value)
{
}
};
Then to destroy a node and all of its children, you'd call reset on the appropriate std::unique_ptr<Node>

Answer 3

I think what you actually want ist this:
struct Node
{
int Value;
Node* _LeftNode;
Node* _RightNode;
Node(int value)
: Value(value), _LeftNode(NULL), _RightNode(NULL)
{
}
~Node() {
delete _LeftNode;
delete _RightNode;
}
};
This way you are using the destructor to clean up recursivly.
delete nullptr is ok btw.
EDIT:
the unique_ptr<> usage in one of the other answers is probably the smarter way to do this.

Answer 4

Given:
struct Node
{
int data = 0;
struct Node * left = nullptr, * right = nullptr;
Node(int data) { this->data = data; }
};
This recursive function deletes a node & its childs (+ one comment):
void DeleteTree(struct Node* node) // A copy of the caller pointer
{
if (node)
{
DeleteTree(node->left); // Recur on left subtree
DeleteTree(node->right); // Recur on right subtree
delete node;
// node = nullptr; <-- This line is useless
}
}
To your wondering "but when I come out the pointer is something else":
There is no point in node = nullptr line, since when you call DeleteTree(my_node) function, node is a copy of my_mode, so when you set node = nullptr it has no effect on my_node that on exit from DeleteTree(my_node) points to a deleted, invalid object.
--
Possible solution:
#define DELETE_TREE(node) DeleteTree(node); node = nullptr; // Macro
int main()
{
struct Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);
DELETE_TREE(root->left); // root->left became nullptr
DELETE_TREE(root); // root became nullptr
return 0;
}
After DeleteTree function, the caller pointer points to an invalid object since its object already released. A possible solution is to define a DELETE_TREE Macro to "auto-nullify" the caller pointer after DeleteTree function.
--
Implementation with Modern C++ Smart Pointers:
#include <memory>
struct Node
{
int data = 0;
std::unique_ptr<Node> left, right;
Node(int data) { this->data = data; }
};
int main()
{
std::unique_ptr<Node> root;
root = std::make_unique<Node>(1);
root->left = std::make_unique<Node>(2);
root->right = std::make_unique<Node>(3);
root->left->left = std::make_unique<Node>(4);
root->left->right = std::make_unique<Node>(5);
root.reset();
return 0;
}