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 > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- ........................ *MOHIT VERMA* -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
