二进制搜索树递归显示功能，不确定如何工作？

Question

I have a BST and I found this function online, which prints it in the correct order, but I don't understand how.

void display()
{
    inOrder(root);
}
void inOrder(treeNode* n)
{
   classA foo;
   if (n != NULL)
   {
        inOrder(n->left);
        foo = *n->item;
        cout << foo << endl << endl;
        inOrder(n->right);
    }
 }

I initialize each node's left and right to NULL. Why does the if statement continue after a NULL pointer is sent in, and where does it continue from?

Answer 1

Say you have the following BST:

When inOrder is called with this BST,

Recursion level 0

n points to 8 . So you enter the block:

    inOrder(n->left);
    foo = *n->item;
    cout << foo << endl << endl;
    inOrder(n->right);

inOrder is called using n->left as argument, which is node 3 .

Recursion level 1

n points to 3 . So you enter the same code block.

inOrder is called using n->left as argument, which is node 1 .

Recursion level 2

n points to 1 . So you enter the same code block.

inOrder is called using n->left as argument, which is NULL .

Recursion level 3

n points to NULL . So you do not enter the above code block. You simply return from the function. Now are back in the next statement in the previous recursion level.

Recursion level 2

The following lines are executed:

    foo = *n->item;
    cout << foo << endl << endl;
    inOrder(n->right);

You print 1 . Then, inOrder is called using n->right as argument, which is NULL .

Recursion level 3

n points to NULL . So you do not enter the above code block. You simply return from the function. Now are back in the next statement in the previous recursion level.

Recursion level 2

There are no more lines to execute. The function simply returns to previous recursion level.

Recursion level 1

The following lines are executed:

    foo = *n->item;
    cout << foo << endl << endl;
    inOrder(n->right);

You print 3 . Then, inOrder is called using n->right as argument, which is node 6 .

You can now follow this step by step all the way. The end result is that you will end up printing the numbers in the following order.

1
3
4
6
7
8
10
13
14