How binary search tree insertion works using recursion? - c++

Question

I'm having some trouble understanding the recursive part of binary search tree insertion.
bstnode* insert(bstnode* root,int data)
{
if(root==NULL){
bstnode* tmp= new bstnode();
tmp->data=data;
tmp->left=tmp->right=NULL;
return tmp;
}
if(data<root->data)
root->left = insert(root->left, data);
else
root->right = insert(root->right, data); //can't understand the logic here
return root;
}
/* consider following BST with their addresses[]:
15 [100]
/ \
10 20 [200]
\
tmp [300]
*/
According to me root->right = insert(root->right, data); should store the address of the newly created node in root->right so this code shouldn't work for tree with height>2.
However, it is working perfectly for any number of nodes.
I must be missing some crucial details here.
suppose I want to insert 25 in BST i.e. insert(root,25);
as 25>15:- I'm breaking down the recursive part here:
root->right = insert(root->right, 25);
or 15->right = insert(15->right,25); Here, recursively calling it again because 25>20
insert(root->right, 25) => root->right->right = insert(root->right->right, 25);
or insert(15->right, 25) => 20->right = insert(20->right, 25);
insert(20->right,25) is NULL so a new node tmp is created.
insert(20->right,25); returns tmp.
unwinding the recursion now.
//20->right = insert(20->right, 25);
so,
20->right= 300 (tmp address);
//insert(15->right, 25) => 20->right
//and 15->right = insert(15->right,25);
15->right = 20->next;
therefore 15->right = [300] address.
or
root->right = [300] address.
what's wrong with my approach?
Again an overview of recursive calls:
15->right = insert(15->right,25);
15->right = [20->right = insert(20->right,25)]; //20->right is NULL so creating new node
15->right = [20->right= 300 address of tmp];
15->right = [20->right or 300]
15->right = [300] // but in reality 15->right = [200]

Answer 1

you are forgetting that root->right is the root->right of the address you are passing into the function as root. every call to insert passes in root->right or root->left depending on which way you traverse.
This statement is incorrect:
root->right = root->right->right = tmp;
once an iteration of the function is returned it is removed from the stack so in this case we have 3 calls I will put your numbers in place of the pointer value.
insert(15->right,25)
insert(20->right,25)
the last one is null so it creates the node with 25 and returns it to the call insert(20->right,25) and sets 25 as 20->right so you have a tree that looks like this
/* consider following BST with their addresses[]:
20 [200]
\
25 [300]
*/
it then returns this tree to the call insert(15->right,25) and sets that trees right to the tree we just returned which so we get your final tree
/* consider following BST with their addresses[]:
15 [100]
/ \
30 20 [200]
\
25 [300]
*/
EDIT: let me see if I can clarify. Lets look at your tree again
/* consider following BST with their addresses[]:
15 [100]
/ \
10 20 [200]
\
tmp [300]
*/
we want to insert 25 so we call (again I will use the value at that node of the tree to represent the pointer we are passing)
insert(15, 25)
this then calls insert on root->right which happens to be 20
insert(20, 25)
this calls insert again on 20 right node now which happens to be null
insert(null,25)
so lets now look at the returns
insert(null,25) returns a node with 25 in it and then is remove from the stack
return 25;
insert(20,25) gets its return of a node with 25. it sets its right child to 25 which looks like this
20->right = 25;
return 20;
now we are back to the original call of insert(15,25). it got returned 20. so it does
15->right = 20;
return 15;

Answer 2

I think the confusion may be coming from two different sources for you.
First the tree commented into your code would not be possible. Second is that a new node is only created when the function is passed in a null pointer. Only values less than 15 can go to the left. It would be something like this instead (depending on add order):
15
/ \
20
/ \
30
When you go to add 25 to this it will end up as follows:
15
/ \
20
/ \
30
/
25
I will try and step through the code on this to explain. When adding 25 to the original tree on the first function call the first node is not NULL and 25 > 15 so the
else
{
root->right = insert(root->right, data);
}
is called. This calls the same insert function recursively but is now using the 20 node as it's comparison. Again not null and 25 > 20 so call insert on right node as above. This again calls the recursive function but now on 30. 25<30 so it calls the function on the left node. At this point the function as been passed in a NULL pointer as there is nothing there and a new node is created and placed in this spot.

Answer 3

Note that insert() always returns the root that was passed to it as an argument unless root == NULL. There's therefore no way for the new node you insert to "walk up the tree". What happens in the recursive call doesn't matter -- you always return the same root that you were passed in the non-NULL case.
Despite the way some people teach recursion, I think it helps (for my brain anyway) to not try to unroll the recursion, and instead consider whether the logic makes sense:
If you are passed a non-NULL node and data < root->data, would you get the correct result if you do root->left = insert(root->left, data) and assume the insert() magically "just works" (i.e., that it inserts data into the left tree and returns the root of that tree)?
If the logic checks out for both the left and right case, you then consider the base case: If you are passed a NULL node, will you return the correct one-element tree?
If the logic checks out for the base case too, then you know your code must be correct, since the recursive steps make sense and you know that you will land in a base case that also makes sense (since you will eventually reach a NULL node as you walk down the tree).

Answer 4

In a way you are correct. You can never have a sub-tree (not tree) of height >2.
In this code, you will never have a root->right->right because, as far as the code is concerned, when you call
root->left = insert(root->left, data);
the (local) root pointer is now pointing to the node you just inserted. the (local) root is pointing to root->left.
Therefore, you CAN have a tree of any height(However, the local root pointer is pointing to a sub-tree of height <2)