Skip to main content

Implementation of binary tree in C++


A binary tree is a non-linear data structure where every node has maximum two branches - left subtree and right subtree. An empty node is also a binary tree.

In earlier posts we have seen how to insert a node to binary tree, how to traverse a tree and how to delete nodes from tree.

The functions look almost similar in C++. Except that all these functions and root pointer are all parts of the class binary tree.

That is a good thing and a bad thing. If root is a data member (so private), how do we use root as a parameter in all the recursive functions - insert, delete, inorder etc?

Easy answer would be don't use recursion. Anyway no one like them.

So we can write insert and delete functions without using recursion and even search function. But traversal functions have to be recursive.

Ok, let us write a function to return root of the tree and use it as first parameter in all traversal functions.

Here is the complete class.

You can download this code along with test program from here


struct node
{
   int num;
   node *left;
   node *right;
};
class binarytree
{
   node *root;
public:
   binarytree();
   node* create_node(int val);
   void insert( int val); 
   void inorder(node *nd);
   void preorder(node *nd);
   void postorder(node *nd);
   node *search(node *nd,int val);
   void delete_node(int val);
   node*find_parent(node *nd);
   node*find_successor(node*nd,node**parent);
   node* get_root();
};
binarytree::binarytree()
{
    root = NULL;
}
node *binarytree::create_node(int val)
{
   node *newn = new node;
   newn->num = val;
   newn->left = NULL;
   newn->right = NULL;
   return newn;
}
void binarytree::insert( int val)
{//insert a node non-recursively
    node*newnode = create_node(val);
    if(root==NULL)
    {//tree is empty
       root = newnode;
       return;
    }
    node *temp = root;
    node *parent = NULL;
    while(temp!=NULL)
    {//find the location for new node
        parent = temp;
        if(temp->num >val)
           temp = temp->left;
        else
           temp = temp->right;
     }
     //which branch
     if(parent->num >val)
        parent->left = newnode;
     else
        parent->right = newnode;    
}

void binarytree::inorder(node *nd)
{
  if(nd!=NULL)
  {
     inorder(nd->left);
     cout<<nd->num<<"  ";
     inorder(nd->right);
   }
}
void binarytree::preorder(node *nd)
{
  if(nd!=NULL)
  {
     cout<<nd->num<<"  ";
     preorder(nd->left);   
     preorder(nd->right);
   }
}
void binarytree::postorder(node *nd)
{
  if(nd!=NULL)
  {
     postorder(nd->left);   
     postorder(nd->right);
     cout<<nd->num<<"  ";
   }
}
node* binarytree::search(node *nd, int val)
{
     if(nd==NULL)
       return nd;
     if(nd->num==val)
        return nd;
     if(nd->num > val)
       return search(nd->left,val);
     if(nd->num <val)
       return search(nd->right,val);
}
node *binarytree::find_parent(node *nd)
{
    if(nd==root)
    //root has no parent
       return NULL;
    node *temp = root;
    while(temp)
    {
       if(temp->left==nd || temp->right==nd)
          return temp;
       if(nd->num > temp->num)
         temp = temp->right;
       else if(nd->num <temp->num)
         temp = temp->left;
    }
    return NULL;
}
node*binarytree::find_successor(node*nd,node**parent)
{//successor is tree minimum of right subtree
   node*temp = nd->right;
   *parent = nd;
   while(temp->left!=NULL)
      {
      *parent=temp;
      temp = temp->left;
      }
   return temp;
}
         
void binarytree::delete_node(int val)
{
    node *dn = search(root,val);
    if(dn==NULL)
      {
        cout<<"value not found\n";
        return;
      }
     node*parent = find_parent(dn);
    if(dn->left!=NULL && dn->right!=NULL)
    {//node has both subtrees. delete successor instead
        node*successor = find_successor(dn,&parent);
        dn->num = successor->num;//copy data
        dn = successor;
    }    
   
    if(dn->left==NULL && dn->right==NULL)
    {//leaf node
       
        if(parent==NULL && dn==root)
           root = NULL;
        else if(parent==NULL)
           cout<<"Error";
        else
           if(dn==parent->left)
              parent->left = NULL;
           else if(dn==parent->right)
              parent->right = NULL;
        delete dn;
    }
    else if(dn->left==NULL||dn->right==NULL)
    {//has one child
         
         node* child = dn->left?dn->left:dn->right;
         if(dn==root)
             root = child;
         else{
            if(parent->left==dn)
              parent->left = child;
            else
             parent->right = child;
          }
          delete dn;
    }
}
node*binarytree::get_root()
{
   return root;
}
Labels:

Comments

Post a Comment

Popular posts from this blog

Program to delete a node from linked list

How do you remove a node from a linked list? If you have to delete a node, first you need to search the node. Then you should find its previous node. Then you should link the previous node to the next node. If node containing 8 has to be deleted, then n1 should be pointing to n2. Looks quite simple. Isn't it? But there are at least two special cases you have to consider. Of course, when the node is not found. If the node is first node of the list viz head. If the node to be deleted is head node, then if you delete, the list would be lost. You should avoid that and make the second node as the head node. So it becomes mandatory that you return the changed head node from the function.   Now let us have a look at the code. #include<stdio.h> #include<stdlib.h> struct node { int data; struct node * next; }; typedef struct node * NODEPTR; NODEPTR create_node ( int value) { NODEPTR temp = (NODEPTR) malloc( size...
3 comments
Read more

Delete a node from doubly linked list

Deletion operation in DLL is simpler when compared to SLL. Because we don't have to go in search of previous node of to-be-deleted node.  Here is how you delete a node Link previous node of node of to-be-deleted to next node. Link next node of node of to-be-deleted to previous node. Free the memory of node of to-be-deleted Simple, isn't it. The code can go like this. prevnode = delnode->prev; nextnode = delnode->next; prevnode->next = nextnode; nextnode->prev = prevnode; free(delnode); And that is it. The node delnode is deleted. But we should always consider boundary conditions. What happens if we are trying to delete the first node or last node? If first node is to be deleted, its previous node is NULL. Hence step 3 should not be used.  And also, once head is deleted, nextnode becomes head . Similarly if last node is to be deleted, nextnode is NULL. Hence step 4 is as strict NO NO. And we should set prevnode to tail. After we put these things together, we have...
Post a Comment
Read more

Deletion of a node from linked list given only that node

We have considered how to create and print the list and delete a node from a list. But often interviewers ask you a question, given a node of the list, how do you delete that node? The problem here is we do not know the head of the list. We just the know one node which we must delete. Well, the solution is much simpler than it appears. Suppose 12 is the node which should be deleted. As we can see from the diagram, we are deleting the next node of 99 by the usual method of linking n to next of next of 12. But the requirement was not to delete next node  but the node with 12 itself. So we will retain the data of that node (99) by copying it to previous node. And now the node with 99 is marked for deletion. Which can be easily deleted as follows. We are assuming that n is the node with 12 as value. n -> data = n -> next -> data; //copy the data n -> next = n -> next -> next; //link to next free( n -> next) ; //delete next of n   ...
Post a Comment
Read more