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# mycodeschool/BSTDeletion_CPP.cpp

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 /* Deleting a node from Binary search tree */ #include using namespace std; struct Node { int data; struct Node *left; struct Node *right; }; //Function to find minimum in a tree. Node* FindMin(Node* root) { while(root->left != NULL) root = root->left; return root; } // Function to search a delete a value from tree. struct Node* Delete(struct Node *root, int data) { if(root == NULL) return root; else if(data < root->data) root->left = Delete(root->left,data); else if (data > root->data) root->right = Delete(root->right,data); // Wohoo... I found you, Get ready to be deleted else { // Case 1: No child if(root->left == NULL && root->right == NULL) { delete root; root = NULL; } //Case 2: One child else if(root->left == NULL) { struct Node *temp = root; root = root->right; delete temp; } else if(root->right == NULL) { struct Node *temp = root; root = root->left; delete temp; } // case 3: 2 children else { struct Node *temp = FindMin(root->right); root->data = temp->data; root->right = Delete(root->right,temp->data); } } return root; } //Function to visit nodes in Inorder void Inorder(Node *root) { if(root == NULL) return; Inorder(root->left); //Visit left subtree printf("%d ",root->data); //Print data Inorder(root->right); // Visit right subtree } // Function to Insert Node in a Binary Search Tree Node* Insert(Node *root,char data) { if(root == NULL) { root = new Node(); root->data = data; root->left = root->right = NULL; } else if(data <= root->data) root->left = Insert(root->left,data); else root->right = Insert(root->right,data); return root; } int main() { /*Code To Test the logic Creating an example tree 5 / \ 3 10 / \ \ 1 4 11 */ Node* root = NULL; root = Insert(root,5); root = Insert(root,10); root = Insert(root,3); root = Insert(root,4); root = Insert(root,1); root = Insert(root,11); // Deleting node with value 5, change this value to test other cases root = Delete(root,5); //Print Nodes in Inorder cout<<"Inorder: "; Inorder(root); cout<<"\n"; }

### meet2mky commented Mar 6, 2018

For everyone who are not getting how the link between the parent of deleted node and and child of deleted node is maintainted. You can easily get it just by looking where the return value of each recursion is going.

### doveral commented Mar 28, 2018

this code is quite very clean to understand,thank you very much ,you did a great job,atleast better than what my lecturer does

### rak96 commented Apr 30, 2018

i have a doubt.. how the ctrl trasfers to else if to else

### vageeesh commented Jul 22, 2018

good explanation.

### Pasha54 commented Sep 29, 2018

In Insert function, data has input as char data type, why?

### sheikhazad commented Dec 21, 2018 • edited

Hi, How are you making sure that once the node to be is deleted is deleted and the link to its parent is still maintained now between the child of the node deleted and its parent ?
More specifically,
else if(root->left == NULL) {
struct Node *temp = root;
root = root->right;
delete temp;
}

After deleting the temp, what about the link between temp's parent and current root ? I am little confused here.
Please explain if I am wrong.
Thanks,
Sidharth

Let me copy paste the funtion with comments so that u may understand:

Node* deleteNode(Node* temp, int data)
{
(temp)? std::cout<<"\nIn deleteNode temp->data = " <data : std::cout<<"\nIn deleteNode temp = NULL" ;
if(temp == NULL)
{
std::cout<<"It's NULL";
return temp;
}
else if( data < temp->data)
{
std::cout<<"\n Data to be delted is in left of : " << temp->data << ", so sending next left node ";
temp->left = deleteNode(temp->left, data); //Assign to temp->left because u are sure now u r not deleting current node but in left of current. temp->left will be parent of what deleteNode() returns when stack unwinds
(temp->left) ? std::cout<<"\nReturned temp->left Node: " <left->data : std::cout<<"\nReturned temp->left = NULL";
}
else if(data > temp->data)
{ std::cout<<"\n Data to be delted is in right of : " << temp->data << ", so sending next right node ";
temp->right = deleteNode(temp->right, data); //Assign to temp->right because u are sure now u r not deleting current node but in right of current. temp->right will be parent of what deleteNode() returns when stack unwinds
(temp->right) ? std::cout<<"\nReturned temp->right Node: " <right->data : std::cout<<"\nReturned temp->right = NULL";
}
else //found data to be deleted
{
Node* current = temp; //for clarity
if(current->left == NULL && current->right == NULL) // If no children
{
delete current;
temp = NULL; // return NULL to be linked to previous parent's left or right in previous "else if" when stack unwinds
std::cout<<"\n No children";
}
//Node to be deleted has left or right or both childred
else if(current->left == NULL)
{
temp = current->right;
delete current; // return current->right to be linked to previous parent's left or right in previous "else if" when stack unwinds, U dont need to track parent in recursion as parent will be there when current recursion unwinds to previous
std::cout<<"\n Has right children";
}
else if(current->right == NULL)
{
temp = current->left;
delete current; // return current->left to be linked to previous parent's left or right in previous "else if" when stack unwinds, U dont need to track parent in recursion as parent will be there when current recursion unwinds to previous
std::cout<<"\n Has left children";
}
else//if left and right children
{
std::cout<<"\n Has left and right children";
Node* minNodeInRight = findMinNode(current->right); //Find min node in right of current node, and that min node will be deleted in future.
std::cout<<"\n Min Node found to be deleted is " << minNodeInRight->data;
current->data = minNodeInRight->data; //Copy min node data here and dont delete current but delete the found node in right of current

``````		//current->right = deleteNode(minNodeInRight->right, minNodeInRight->data); // U cant pass minNodeInRight->right as argument becoz its much below current node, if u send it link with parent will be broken,
//U need to traverse from Current node until u find  minNodeInRight->data
current->right = deleteNode(current->right, minNodeInRight->data);	//Assign to current->right because u are sure now u r not deleting current node but in right of current.

}

}
(temp) ? std::cout<<"\n END OF DELTE, returning Node: " <<temp->data : std::cout<<"\nReturning temp = NULL";
return temp;
``````

}

### hardikpatel7 commented Sep 17, 2019

What need of temp?
Isn't it enough
root=root->left
for the case 2

for making memory free . because that data is not useful for us now. but it is stored in Heap section.

### himu59 commented Oct 31, 2019 • edited

why we assign delete function to root in line 87.
it is necessary?

### paras062 commented Jan 29, 2020

Looks like it fail if you try to delete 12 from the BST, as 12 is equal to root.data, and we don't have condition to handle when data = root.data.
We have condition to check if data > root.data or data < root.data, what missing is data = root.data

I tried this on my end, below is the python version of doing the same.

# CASE 1: Delete root of the binary tree

``````elif (data == root.data):
temp = root
newRoot = root.left
root = root.left

# Move current root to the end of right most node of the left subtree
while(root.right != None):
root = root.right

if root.right == None:
root.right = temp.right

del temp, root
return newRoot
``````

### paras062 commented Jan 29, 2020

There is another issue with the code, in your example you've been deleting elements from the right sub tree, if you try to delete from left subtree, your FindMin function will fail.

def minValue(node, data):
current = node
if (data < node.data):
# loop down to find the right most leaf
while(current.right is not None):
current = current.right
elif (data > node.data):
# loop down to find the left most leaf
while(current.left is not None):
current = current.left

``````return current.data
``````

This is what I did to fix it.

### pnguptacs7 commented Feb 24, 2020

Hi paras062,
we don't need to make that change.The code is similar to search element in a binary tree ie element can be in a left sub tree or right sub tree,all FindMin function does is to find min element in a right sub tree.I executed this code and it works fine.

### TheGupta2012 commented Mar 24, 2020

Hey this is Harshit here.
I just wanted to know if we could directly update the data of root by making a function findmin which returns the integral value of the minimum element of the right subtree. I tried to do it but my final output after deleting some element is coming out exactly the same.
Would appreciate it if anyone could tell me what is wrong in the code;
MY UPDATED CODES.
else // the updated case.
{
root->data=findmin(root->right);
root->right=deleteNode(root->right, root->data);
return root;
}
int findmin(bnode *root) //the findmin function
{
if(root==NULL)
return -1;
else
{
bnode *temp;
temp=root;
while(temp->left!=NULL)
temp=temp->left;
return temp->data;
}
}

### elakashghosh commented Apr 17, 2020

What need of temp?
Isn't it enough
root=root->left
for the case 2

you have to free the memory

### vaibhavs017 commented Jul 12, 2020

shouldn't the second case be like this :
else if(root->right==NULL){
Node* temp=root->left;
delete root;
root=temp;
}

You must delete the temp as it is holding the address of the root in the memory. And at the last delete the root for deletion of the allocation of the root node in the heap

### vaibhavs017 commented Jul 12, 2020 • edited

What need of temp?
Isn't it enough
root=root->left
for the case 2

you have to free the memory

As root is the local variable this might give error. So it is better to store the address in temporary variable and then delete it.

### vaibhavs017 commented Jul 12, 2020

why we assign delete function to root in line 87.
it is necessary?

Yes the memory allocate in the heap section must be free. If not deleted then this may lead to program crash or memory leak

### pavan-areti commented Aug 22, 2020

What need of temp?
Isn't it enough
root=root->left
for the case 2

root = root.left will just give attachment with its left pointer but here temp is referring to a minimum in the right subtree of the root;

### ghoshdipan commented Oct 3, 2020

shouldn't the second case be like this :
else if(root->right==NULL){
Node* temp=root->left;
delete root;
root=temp;
}

I think this is good as it seems logical to me. I am not clear with the source code.
And what will be the code of findMin??

### sanu-coder commented Oct 26, 2020

Very clean and clear code. Thanks..

### rahuldeepattri commented Jan 19, 2021 • edited

A small bug, the Case 2 logic should be like this (ignoring delete statements):

``````             //Case 1 node has no children
if (root.left == null && root.right == null) {
root = null;
}
// Case 2 Only one child
else if (root.left != null && root.right == null) {
root = root.left;
} else if (root.right != null && root.left == null) {
root = root.right;
}
//Case 3 Both the child are present
else {
``````

The original logic is <3

### ajaypep commented Jan 27, 2021 • edited

I can't seem to delete the root node, if I did it breaks the link in the right sub tree.

```#include<iostream>
#include<sstream>
struct node
{
int data;
node *left,*right;
};

class bst
{
node *root;
public:
bst()
{
root=nullptr;
}
{
if(root==nullptr)
{
root=new node;
root->data=info;
root->left=nullptr;
root->right=nullptr;
std::cout<<"Inserted "<<root->data<<"\n";
return root;
}
else if(info==root->data)
{
return root;
}
else if(info<root->data)
{
}
else
{
}
return root;
}
void insert()
{
std::string s;
int i;
do
{
std::cout<<"Enter q to quit\n";
std::cout<<"Enter a unique number : ";
std::cin>>s;
if(s=="q")
{
std::cout<<"Exited\n";
return;
}
std::stringstream int_str(s);
int_str>>i;
}while(true);
}
void look(node* root,int info)
{
if(root==nullptr)
{
}
else if(info==root->data)
{
std::cout<<"Node found containing "<<info<<"\n";
}
else if(info<root->data)
{
look(root->left,info);
}
else
{
look(root->right,info);
}
}
void search()
{
int i;
std::string s;
do
{
std::cout<<"Enter q to quit\n";
std::cout<<"Enter a number to search  : ";
std::cin>>s;
if(s=="q")
{
std::cout<<"Exited\n";
return;
}
std::stringstream int_str(s);
int_str>>i;
look(root,i);
}while(true);
}
node* findMin(node *root)
{
while(root->left!=nullptr)
{
root=root->left;
}
return root;
}
node* remove(node* root,int info)
{
//first search the node
if(root==nullptr)
{
return root;
}
else if(info<root->data)
{
root->left=remove(root->left,info);
}
else if(info>root->data)
{
root->right=remove(root->left,info);
}
//found the element
else
{
//it has no child
if(root->left==nullptr && root->right==nullptr)
{
delete root;
std::cout<<"Deleted "<<root->data<<"\n";
return root;
}
//it has one child(right)
else if(root->left==nullptr)
{
node* tmp=root->right;
delete root;
std::cout<<"Deleted "<<root->data<<"\n";
return tmp;
}
//it has one child(left)
else if(root->right==nullptr)
{
node* tmp=root->left;
delete root;
std::cout<<"Deleted "<<root->data<<"\n";
return tmp;
}
//it has two children
else
{
node* tmp=findMin(root);
root->data=tmp->data;
root->right=remove(root->right,tmp->data);
}
}
return root;
}
void del()
{
int i;
std::string s;
do
{
std::cout<<"Enter q to quit\n";
std::cout<<"Enter a number = ";
std::cin>>s;
if(s=="q")
{
std::cout<<"Exited\n";
return;
}
std::stringstream int_str(s);
int_str>>i;
root=remove(root,i);
}while(true);
}
};
int main()
{
bst myt;
myt.insert();
myt.search();
myt.del();
myt.search();
return 0;
}```

### mmatros commented Apr 24, 2021 • edited

@snake-jazz

``````delete root;
std::cout<<"Deleted "<<root->data<<"\n";
``````

### ajaypep commented May 6, 2021

@snake-jazz

``````delete root;
std::cout<<"Deleted "<<root->data<<"\n";
``````

Hey, thanks for replying to my post. Actually I've moved to Java so I'm practicing data structures using Java, still appreciate it.

### Bhabesh142 commented Jun 6, 2021 • edited

Hello All, why do we need to

Node* FindMin(Node* root){ while(root->left != NULL) root = root->left; return root;}

find Min and use of Inorder here when we have already used it in delete Function.? I have a doubt on this.?
can anybody help me here.? And one more mistake is char data is used instead of integer(int data). so kindly change it.?

### jankivaghasiya commented Sep 13, 2021

What need of temp?
Isn't it enough
root=root->left
for the case 2

There is another issue with the code, in your example you've been deleting elements from the right sub tree, if you try to delete from left subtree, your FindMin function will fail.

def minValue(node, data):
current = node
if (data < node.data):

# loop down to find the right most leaf

while(current.right is not None):
current = current.right
elif (data > node.data):

# loop down to find the left most leaf

while(current.left is not None):
current = current.left

``````return current.data
``````

This is what I did to fix it.

We need to FindMax in order to delete from left subtree as Max from left subtree would be Inorder Predecesor

### phongvu009 commented Nov 18, 2021

Hello All, why do we need to

Node* FindMin(Node* root){ while(root->left != NULL) root = root->left; return root;}
find Min and use of Inorder here when we have already used it in delete Function.? I have a doubt on this.? can anybody help me here.? And one more mistake is char data is used instead of integer(int data). so kindly change it.?

Inorder function to print the tree in order list like this : 1 3 4 5 10 11

### AryanGitHub commented Feb 5, 2022 • edited

in the code given by mycodeschool, there is a logical bug, which can cause a Segmentation Error
`root=NULL`, and `root = root->right;` this will not be reflected outside the function,
it should be `*root=NULL`, and the function definition should be like this `struct Node* Delete(struct Node **root, int data) `

### Akash2791-1 commented Apr 1, 2022

my approach without recursive (gfg practice):-

void delLeafnode(Node *cur, Node *prev, int X)
{
if(Xdata) prev->left = NULL;
else prev->right = NULL;
}

void deloneNode(Node *cur, Node *prev, int X)
{
if(Xdata)
{
if(cur->left) prev->left = cur->left;
else prev->left = cur->right;
}
else
{
if(cur->left) prev->right = cur->left;
else prev->right = cur->right;
}

}

Node *findmin(Node *root, int X)
{
Node *cur = root->right;
Node *prev = root;
Node *temp;
while(cur!=NULL)
{
if(!cur->left)
{
temp = cur;
if(!cur->left && !cur->right) delLeafnode(cur, prev, X) ;
else if(!cur->left || !cur->right) deloneNode(cur, prev, X);
}
prev = cur;
cur = cur->left;
}
return temp;
}

Node *deleteNode(Node *root, int X) {
Node *cur = root;
Node *prev = NULL;

``````if(root->data==X && (!root || (!root->left && !root->right))) return NULL;
while(cur!=NULL)
{
if(cur->data == X)
{
if(!cur->left && !cur->right)
{
delLeafnode(cur, prev, X);
}

else if(!cur->left || !cur->right)
{
deloneNode(cur, prev, X);
}

else if(cur->left && cur->right)
{
Node *node = findmin(cur, X);
cur->data = node->data;
}
}

prev = cur;
if(X<cur->data) cur = cur->left;
else cur = cur->right;
}
return root;
``````

}

### vinamrgrover commented Apr 14, 2022 • edited

in the code given by mycodeschool, there is a logical bug, which can cause a Segmentation Error
`root=NULL`, and `root = root->right;` this will not be reflected outside the function,
it should be `*root=NULL`, and the function definition should be like this `struct Node* Delete(struct Node **root, int data)`

How this will not be reflected outside the function if he has equated the function to the original root variable by `root = Delete(root,5);`?
By using pointer to a pointer you are creating a mess for yourself, nothing else.
Better check this with the Search_Node function.

### vinamrgrover commented Apr 14, 2022 • edited

Hey, I just want everyone to know that after Deleting a node from a BST, better check if the node is present or not by the `Search_Node` Function. Call the `Search_Node` function in the main function and check if the value returned by the `Search_Node` function is `true` or `false.`
Here is the Definition:
bool Search_Node(BstNode *root, int data)
{
if(root==NULL)
return false;
if(data==root->data)
return true;
else if(data>root->data)
return Search_Node(root->right, data);
else
return Search_Node(root->left, data);
}

Here's how you can check:
if(Search_Node(root, 20)==true) // here 20 is the data for the deleted node
printf("FOUND\n");
else