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Inorder binary search tree

Hi,

I am trying to understand recursion for inorder traversal of : http://geeksquiz.com/binary-search-tree-set-2-delete/

// A utility function to do inorder traversal of BST
void inorder(struct node *root)
{
if (root != NULL)
{
inorder(root->left);
printf("%d ", root->key);
inorder(root->right);
}
}

Say the tree is the following...

/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */

For inorder, we get 20,30,40,50, 60,70,80

Am i correct in understanding the recursion code is that, until you have not reached the most left, just keep on going left... is this right? any help to get me started is really appreciated.
if (root != NULL)
{
inorder(root->left);

Guy Hengel [angelIII / a3]

yes, this is correct understanding
the "output (printf function here)" is after the "inorder" function, so you need to proceed to get the leftmost node, then print that node (as there both left and right are null then, and the called function is not doing anything anymore)

zizi21

ASKER

Thanks for your reply.

But, if both left and right are null, this does not get executed if (root != NULL)
sorry, if this sounds stupid but how does it get printed then..

ASKER CERTIFIED SOLUTION

Guy Hengel [angelIII / a3]

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zizi21

ASKER

Thanks for replying.

Am I right, that big oh notation for this is O(n) ?

SOLUTION

pepr

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