BST中的顺序遍历

Question

I have a balanced binary search tree of integers and I want to find the leftmost node which stores the integer greater or equal to a fixed number like a using a function like ask(a) .

for example suppose that I have added the following points in my tree, 8,10,3,6,1,4,7,14,13

Then the tree would be like this:

now ask(1) should be 1 , ask(3) should be 3 , ask(2) should be 3 and so on.

I think that I can use Inorder traversal to write my ask function, But I don't know how.

Iv written this piece of code so far:

inorderFind(node->left, a);
if (node->key.getX() >= a)
    return node;
inorderFind(node->right, a);

The first argument is the current tree node and a is the a that is described above. I know that I can use a bool variable like flag and set it to true when the if condition holds, and then it would prevent from walking through other nodes of the tree and returning a false node. Is there anything else that I can do?

Answer 1

Trees have the wonderful property of allowing queries through simple, recursive algorithms. So, let's try to find a recursive formulation of your query.

Say LEFTMOST(u) is a function which answers this question :

Given the binary search subtree rooted at node u , with(possibly null) left and right children l and r , respectively, what is the left-most node with a value >= a ?

The relation is quite simple:

LEFTMOST(u) = LEFTMOST(l) if it exists
              LEFTMOST(r) otherwise

That's it. How you translate this to your problem and how you handle concepts like "null" and "does not exist" is a function of your representation.