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Binary tree deletion

Do not get all scared and worried. It is not rocket science (or as I would like to call it - it is not regex).

Remember deleting a node from linked list. When you deleted a node, you did the following 2 steps
  1. Free the memory of the node
  2. Link the previous node of the deleted node to the next node

You will have to do these things in binary tree too. But the difficulty here is do you link the previous node - or parent node in tree terminology, to the left child of deleted node? Or to the right child? You can not link both children, as the parent already may have one child node.

So before we solve this, let us categorize the deletion with the help of a diagram.



Consider the cases

  1. The node to be deleted is a leaf node. That is, it does not have left or right child. In this case, link the parent to NULL in place of deleted node. 
    • If you want to delete 13 - which is a leaf node, you set 14->left to NULL and free memory of node 13.
    • If you want to delete 7 - which is also a leaf node and is the right child of 6, you set 6->right to NULL and free 6.
  2. The node is to deleted is having only one subtree and hence one child node. In this case link the parent to this child node.
    • If you want to delete 14 - which has only left child node, you link the only child of 14 to the parent. That is you set 10->right to 14->left. Now 10 will have right child as 13. And BST property holds good.
    • Instead if you want to delete 10 which has only right subtree - 14 and 13, you link parent of 10 to  14. That is, you set 8->right to 14. 
  3. The node to be deleted has both branches. In this case, you find the successor of the node - which will be treeminimum of right sub tree. Copy the data of this tree minimum to the node to be deleted. And finally delete this tree minimum.
    • If you want to delete 3, you search for minimum of right subtree of 3. Right subtree of 3 has 3 nodes in it, 6,4 and 7. Minimum of this is 4. Copy the data - 4 into the node to be deleted -which is 3. Now you delete the node with 4.  
Delete the node with only left subtree

Delete the node with only right subtree

Delete the node with both subtrees

 Why tree minimum of right subtree?

When you delete a node, its place can be taken by its predecessor or its successor. We are choosing successor here is a substitute for node to be deleted. And we are preserving its information by copying it into the original to-be deleted node. 

Again, why not any other node? Because this subtree minimum will be either a leaf node or it will have only right branch. So we are reducing it to case 2. - that is node with only one branch or 0 branches. 

Next let us look at the code. I have written a recursive function here. You can also a find function to find tree minimum.

The advantage of recursive function is, you don't have to write separate functions to search a node, find the parent node etc.  Of course at the expense of simplicity.

NODEPTR delete_node(NODEPTR nd,int delval)
{
 if(nd==NULL)
          return nd;
 if(nd->val >delval)
           nd->left = delete_node(nd->left,delval);
        else if(nd->val < delval)
           nd->right = delete_node(nd->right,delval);
        else/*node found*/
        {
           if(nd->left==NULL && nd->right==NULL)/*leaf node*/
               {
                   free(nd);
     nd = NULL;
               }
           else if(nd->left==NULL)/*only right child*/
              {
                 NODEPTR temp = nd->right;
                 free(nd);
                 nd =temp;
              }
           else if(nd->right==NULL)/*only left child*/
              {
                 NODEPTR temp = nd->left;
                 free(nd);
                 nd = temp;
              }
           else/*both branches*/
            {
              NODEPTR min_node = find_rightst_min(nd->right);
              nd->val = min_node->val;/*copy value of minimumNode*/
              nd->right = delete_node(nd->right,min_node->val);/*delete minimumNode*/
            }         
      }
    return nd;  
}

/*go left till you reach null*/
NODEPTR find_rightst_min(NODEPTR nd)
{
     NODEPTR temp = nd;
     while(nd)
     {
        temp = nd;
        nd = nd->left;
      }
      return temp;
}

Now let us write the complete program where you keep deleting different nodes in a loop.


#include<stdio.h>
#include<stdlib.h>
struct node
{
   int val;
   struct node *left;
   struct node *right;
};
typedef struct node *NODEPTR;

NODEPTR create_node(int num)
{
     NODEPTR temp = (NODEPTR)malloc(sizeof(struct node));
     temp->val = num;
     temp->left = NULL;
     temp->right = NULL;
     return temp;
}

NODEPTR insert_node(NODEPTR nd,NODEPTR newnode)
{
    if(nd==NULL)
       return newnode;/* newnode becomes root of tree*/
    if(newnode->val > nd->val)
        nd->right = insert_node(nd->right,newnode);
    else if(newnode->val <  nd->val)
        nd->left = insert_node(nd->left,newnode); 
    return nd;   
}

void in_order(NODEPTR nd)
{
   if(nd!=NULL)
    {
        in_order(nd->left);
        printf("%d---",nd->val);
        in_order(nd->right);
     }
}

void pre_order(NODEPTR nd)
{
   if(nd!=NULL)
    {
        printf("%d---",nd->val);   
        pre_order(nd->left);
        pre_order(nd->right);
            
     }
}

void post_order(NODEPTR nd)
{
   if(nd!=NULL)
    {    
 post_order(nd->left);
        post_order(nd->right);
        printf("%d---",nd->val);
     }
}


/*go left till you reach null*/
NODEPTR find_rightst_min(NODEPTR nd)
{
     NODEPTR temp = nd;
     while(nd)
     {
        temp = nd;
        nd = nd->left;
      }
      return temp;
}
NODEPTR delete_node(NODEPTR nd,int delval)
{
 if(nd==NULL)
          return nd;
 if(nd->val >delval)
           nd->left = delete_node(nd->left,delval);
        else if(nd->val < delval)
           nd->right = delete_node(nd->right,delval);
        else/*node found*/
        {
           if(nd->left==NULL && nd->right==NULL)
               {
                   free(nd);
     nd = NULL;
               }
           else if(nd->left==NULL)/*leaf node*/
              {
                 NODEPTR temp = nd->right;
                 free(nd);
                 nd =temp;
              }
           else if(nd->right==NULL)/*leaf node*/
              {
                 NODEPTR temp = nd->left;
                 free(nd);
                 nd = temp;
              }
           else
            {
              NODEPTR min_node = find_rightst_min(nd->right);
              nd->val = min_node->val;
              nd->right = delete_node(nd->right,min_node->val);
            }
         
      }
return nd;
 
  
}
int main()
{
       NODEPTR root=NULL,delnode; 
       int n;
       do
       {
           NODEPTR newnode;
           printf("Enter value of node(-1 to exit):");
           scanf("%d",&n);
           if(n!=-1)
            {  
               newnode = create_node(n);
               root = insert_node(root,newnode);
             }
       } while (n!=-1);
       
       printf("\nInorder traversal\n");
       in_order(root);
      
       
       while(1){ 
   printf("Enter node to be deleted(-1 to stop)");
         scanf("%d",&n);
  if(n==-1)
   break;
           root = delete_node(root,n);
           printf("now tree is");
  in_order(root);
       }    
       return 0;
}
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