mohit, wat if the tree is growing dynamically?

On Sun, Aug 28, 2011 at 11:27 PM, mohit verma <[email protected]> wrote:

> if the number of levels is known - while traversing the tree in BFS order
> keep a loop to print spaces in number- n/2,n/2-1,n/2-2 and so on ,before
> entering at each level. Now if you find any child node empty just print a
> blank space in place of its value.
>
>
> On Sun, Aug 28, 2011 at 10:01 PM, Dave <[email protected]> wrote:
>
>> @Navneet: I suggest that you do an in-order traversal. Assign x values
>> to the nodes sequentially, with y values based on the depth from the
>> root. Thus, in your example, d has coordinates (0,2), b: (1,1), e:
>> (2,2), a: (3,0), f: (4,2), c: (5,1), g: (6,2).
>>
>> Dave
>>
>> On Aug 28, 9:46 am, Navneet Gupta <[email protected]> wrote:
>> > Hope the question is clear. Basically you need to print a given tree
>> > such that spaces will depict the left/right relation at every level.
>> >
>> > output should be something like
>> >                     a
>> >                  b     c
>> >               d   e  f   g
>> >
>> > Levels are separated by new lines. Notice that space between nodes at
>> > higher levels increases with the number of levels we have. Assume a
>> > max of 10 levels. But the algorithm should scale.
>> >
>> > --
>> > Regards,
>> > Navneet
>>
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>
>
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> ........................
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