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# Binary tree output to screen

Posted on 2007-11-29
Medium Priority
1,085 Views
I'm trying to create a function that prints the binary tree out to the screen  making each column start 4 spaces to the right of the precious column.  Here is what I have so far.  I seem to have something out of order though.  Any thoughts ?

``````template <class ItemType>
void RTreeType<ItemType>::ScreenPrint (std::ostream& outFile, NodeType* tree, int spaces) const
{
if ( tree != NULL )
{
ScreenPrint(outFile,tree->right,spaces + 4);
for ( int x=1; x <= spaces; ++x)
cout << setw(4) << " ";
cout << tree->info << endl;
ScreenPrint(outFile,tree->left, spaces + 4);
}
}
``````
0
Question by:PMG76
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LVL 40

Expert Comment

ID: 20377587
It would help if you expanded a bit on what your actual problem is. Whilst it may very well be obvious to you it is not so obvious to others.
0

LVL 53

Expert Comment

ID: 20377634
If I understand it correctly, you want to output the tree like this, correct ?

1
2       3
4   5   6   7

If so, that's quite hard to do with a recursive function.
0

LVL 53

Expert Comment

ID: 20377659
One approach would be to get the depth of the tree (to calculate the amount of spaces needed).

Then print one level of the tree at a time. The spaces at the beginning and between the values can be calculated based on the current level and the total depth of the tree.
0

Author Comment

ID: 20377672
yes that is what I need.  I'm aware of the difficulty, but I have to use recursion.  Right now my out put is like this:

999
500
123
2

So something is not working properly
0

LVL 53

Accepted Solution

Infinity08 earned 1000 total points
ID: 20377736
>> I'm aware of the difficulty, but I have to use recursion.

Here's a way that is close to what I described earlier. Call the function recursively, rising in the tree starting from the lowest level, until you get to the root (highest level). Using the same calculation of amount of spaces, you can then print the current level. Something like :

void printTree(TreeNode node, int level) {
printTree(node->father, level + 1);
// calculate amount of spaces based on current level (level starts counting from the bottom of the tree)
// print the current level, using node->sibling to get the next sibling
}

call by passing the leftmost node on the lowest level as the first parameter, and 0 as second parameter.
0

LVL 29

Expert Comment

ID: 20380428
If I understand the homework assingment well, one text line should display one node (tree rotated 90 degrees). If this is the case then the code is almost O.K. Only the for loop for generating the spaces can be simplified by using the variant of the std::string constructor that takes the counter and the character (remove the for loop completely.

By the way, your output can be fine if the root node contains 2, the right subtree has 999 in the top node, the left subtree is empty, etc.
``````cout << std::string(spaces, ' ') << tree->info << endl;
``````
0

LVL 39

Assisted Solution

itsmeandnobodyelse earned 1000 total points
ID: 20383286
>>>> I'm aware of the difficulty, but I have to use recursion.

You could fill arrays instead of printing. And print the arrays with an outer function

template <class ItemType>
void RTreeType<ItemType>::ScreenPrint ()
{
std::vector<std::vector<ItemType> > out;
// now call the below recursive function with root node and level 0
...

// here you need to iterate the arrays level by level and print

}

template <class ItemType>
void RTreeType<ItemType>::getValues(std::vector<std::vector<ItemType> >& out, NodeType* tree, int level) const
{
if ( tree != NULL )
{
if ((int)out.size() <= level)
out.push_back(std::vector<ItemType>());  // add a new vector if new level
out[level].push_back(tree->info );
ScreenPrint(out, tree->left,level+1);
ScreenPrint(outFile,tree->right, level+1);
}
}

That should bring you further but you still have to handle the spaces somehow. Maybe, by using a string instead of the inner vector<ItemType>. Then you have to insert spaces when an empty node occurs. And have to calculate offsets for each higher level.

You might consider to print the tree turned by 90 degrees:

2

123
500
999

While the depth of a (somehow balanced) tree hardly was getting big, the width of a tree was growing exponentially.

Regards, Alex
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