The tree appears to wind up like this:

10

/ \

9 8

/ / \

2 7 (6)

/ \

3 5

I'm not sure where the 6 goes. The phrase S.RightSubTree().AttachLeft(6,success) appears to plunk it ontop of the 7. For the sake of continueing I'll assume there was a typo there and that it should have been AttachRight.

Now, the traversal of a tree involves checking if the current node is a leaf, traversing the left subtree, then the right subtree. The difference between pre-order, in-order and post-order lies in when you access/print/show the data. In post-order the data is shown last. So, first we go down the left subtree. That repeats till we get to 3. This is a leaf, traversing the left and right subtree produces nothing so then 3 is shown. Now we go back up to 2. We now first have to traverse its right subtree, which is a leaf so we show 5. Now we show 2 (having done its left and right subtrees). Then back up to 9 (with no right subtree) then to 10. 10 cannot be shown yet, first we have to traverse its right subtree, this leads us down to 7, then 6, then 8 then 10. So assuming I have the 6 in the right spot the answer is:

3, 5, 2, 9, 7, 6, 8, 10

The answer you showed looks like the in-order traversal of the tree, where the data is shown between the traversal of the left and right subtrees (which would put 6 in the spot I assumed).

As for the next question, the tree R would appear to be:

20

/ \

1 30

and I'm guessing that the statement binTreeClass T(4,R,S) means that T is a new tree with a root node of 4, a left subtree consisting of R and a right subtree of S. If that's correct a post-order traverse would be:

1, 30, 20, 3, 5, 2, 9, 7, 6, 8, 10, 4

How are my guesses holding up this time?

10

/ \

9 8

/ / \

2 7 (6)

/ \

3 5

I'm not sure where the 6 goes. The phrase S.RightSubTree().AttachLef

Now, the traversal of a tree involves checking if the current node is a leaf, traversing the left subtree, then the right subtree. The difference between pre-order, in-order and post-order lies in when you access/print/show the data. In post-order the data is shown last. So, first we go down the left subtree. That repeats till we get to 3. This is a leaf, traversing the left and right subtree produces nothing so then 3 is shown. Now we go back up to 2. We now first have to traverse its right subtree, which is a leaf so we show 5. Now we show 2 (having done its left and right subtrees). Then back up to 9 (with no right subtree) then to 10. 10 cannot be shown yet, first we have to traverse its right subtree, this leads us down to 7, then 6, then 8 then 10. So assuming I have the 6 in the right spot the answer is:

3, 5, 2, 9, 7, 6, 8, 10

The answer you showed looks like the in-order traversal of the tree, where the data is shown between the traversal of the left and right subtrees (which would put 6 in the spot I assumed).

As for the next question, the tree R would appear to be:

20

/ \

1 30

and I'm guessing that the statement binTreeClass T(4,R,S) means that T is a new tree with a root node of 4, a left subtree consisting of R and a right subtree of S. If that's correct a post-order traverse would be:

1, 30, 20, 3, 5, 2, 9, 7, 6, 8, 10, 4

How are my guesses holding up this time?