adding a node to a Binary Tree from left to right

I have no idea what you just asked.

Ok... let me try again.

In my binary tree, I want to add a node and have it automatically populate the tree from left to right. Then go down a level, again add itself to the tree from left to right.
so...

       a                      a   /   \                  /    b      c   then         b     c                        /                       d

and so forth.

Clearer?

Ok. Give your tree an extra bit of information: each node keeps track of the minimum depth of each of its sub trees. If a node has no subtrees its depth is 0. If it has two subtrees its depth is min(depth(left), depth(right)). Can you see how you could use that information to decide where to add the next node?

ok... so checkDepth(), if a node has a depth of 0, add a left node. then checkDepth() again, if depth != 0 add a right node.

i take it, this will be a recursive task.

Not really. That would cause you to create a tree where you build on the right with only one left node along the spine. Think of depth as a balancing factor. Also, it can be done iteratively.

It seems like the grasshopper will be in the temple a little longer than intended.

I foresee a complication if the subtrees have equal length.

Here is an idea I'll throw out.

// assume node is a pointer typevoid add(node tree, node newbie) { list<node> nodelist int depth = 1 list.pushback(tree)  while( !add_node (nodelist, newbie, depth) );}bool add_node (list&, node, depth&) { for i from 0 to depth:   temp = list.popfront()   if either temp.child == null:      that temp.child = node      return true   else      list.pushback(temp.leftchild)      list.pushback(temp.rightchild)   depth = depth*2}

Basically scan down the tree one level at a time, keeping track of all the nodes on a horizontal level, and adding the node to the first null encountered from left to right.

No, there are no complications if the two subtrees have equal depth. Draw out a tree where the two subtrees have equal depth that has been created by this type of algorithm. Do you see where the node would go?

I'm not gonna lie to you... I don't get the code you just posted.

I see a problem finding out how to increment the mindepth counter on each node, because if the two subtrees are equal in length, then the depth counter should not change. If they are unequal we increment the depth counter.

Oops, depth should be min(depth(left), depth(right)) + 1, and 0 if there are no sub trees. Make more sense now?

Let's pretend you are the recursive function.

I pass you the current node and the node to add. You see that:
curr.depth = 8
curr.left.depth = 8
curr.right.depth = 7

Do you increment the minimum depth counter?

You would add the node to the appropriate subtree. Doing so would give you sufficient information as to whether or not the increment the depth or not.

Quote:

Original post by Alpha_ProgDes now I'm talking about plain ol' Binary Trees, not Binary Search/Sort Trees. I wanted to know (without using arrays as my base) if there is a way to add a node and have it insert and fill the tree from left to right.

You can do the same thing that would happen when using an array, but actually use a binary tree.
All you need to do is keep track of the current number of items in the tree, which you probably do anyway. Then with that number, each bit tells you whether you go left or right during insertion.
e.g. A tree with 5 items:

      A  B       CD   E

First increment the count. This gives us 6. 6 is 110 in binary. Now 0 means go left, 1 means go right. Ignore the most significant digit, and start from the next most significant bit, proceeding towards the lest significant bit.
So you go right, then insert on the left.

Now that you have 6 items, add 1 to that and get 7. 7 is 111 in binary. So you go right, then insert on the right.
Now you have 7 items. +1 gives 8. 8 = 1000. So you go left twice and insert on the left. etc...

No need for recursion, or storing data in each node, or any inefficient algorithm.[cool]

[sad]

How can I become leet like you iMalc?

I was just thinking that it is so easy for a heap to do exactly this with no extra storage, so there has to be a way to do it with a tree. Then I just got lucky I suppose.

Quote:

Original post by iMalc Quote: Original post by Alpha_ProgDes now I'm talking about plain ol' Binary Trees, not Binary Search/Sort Trees. I wanted to know (without using arrays as my base) if there is a way to add a node and have it insert and fill the tree from left to right. You can do the same thing that would happen when using an array, but actually use a binary tree. All you need to do is keep track of the current number of items in the tree, which you probably do anyway. Then with that number, each bit tells you whether you go left or right during insertion. e.g. A tree with 5 items: A B CD E First increment the count. This gives us 6. 6 is 110 in binary. Now 0 means go left, 1 means go right. Ignore the most significant digit, and start from the next most significant bit, proceeding towards the lest significant bit. So you go right, then insert on the left. Now that you have 6 items, add 1 to that and get 7. 7 is 111 in binary. So you go right, then insert on the right. Now you have 7 items. +1 gives 8. 8 = 1000. So you go left twice and insert on the left. etc... No need for recursion, or storing data in each node, or any inefficient algorithm.[cool]

Dude!!! That's #@$%ing genius!!! You sir are a genius.
Now I'm still trying to figure out the mindepth thing.

yeah uh that minimum depth algorithm .... my brain hits wall.

If you use iMalc's algorithm, then you don't need to keep track of mindepth.

But here is how I understood the insert if you keep track of mindepth.

// untested!// returns mindepthint insert(node Node, node NewNode) {  if (Node.left = NULL)    Node.left = NewNode;    return 0;  else if (Node.right == NULL)    Node.right = NewNode;    return 1;  if (Node.left.mindepth <= Node.right.mindepth)    Node.mindepth = min ( Node.right.mindepth, insert(Node.left, NewNode) );  else    Node.mindepth = min ( Node.left.mindepth, insert(Node.right, NewNode) );  return Node.mindepth;}