I've have one solution using two stacks, fs (forward stack) to print
elements in forward manner and rs to print in reverse.
Basic idea is while printing elements in forward order, put the elements in
reverse stack so that the while popping the reverse stack the last element
in row is popped first. Similary while popping reverse stack, make sure, the
first element in the next level is entered last in the stack for the next
iteration of forward printing.
if(root)
fs.push(root)
while (!fs.empty() || !rs.empty()){
while(!fs.empty()){
elem = fs.pop();
print elem.data;
if(elem-left) rs.push(elem-left);
if(elem-right) rs.push(elem-right);
}
while(!rs.empty()){
elem = rs.pop();
print elem.data;
if(elem-right) fs.push(elem-right);
if(elem-left) fs.push(elem-left);
}
}
On Mon, Nov 23, 2009 at 10:47 AM, Nayn nayanish.hi...@gmail.com wrote:
Honestly.. guys.. i would appreciate if we could come up with
something concrete.
On Nov 20, 5:38 pm, ganesa thandavam gthanda...@gmail.com wrote:
we need to use queue and stack alternately ...
once we handle this , i think it should be straight forward to
code ...
On Nov 20, 9:27 am, dinesh bansal bansal...@gmail.com wrote:
On Wed, Nov 18, 2009 at 8:05 AM, Nayn nayanish.hi...@gmail.com
wrote:
Hi guys,
Recently I came across a problem. We've to display a binary tree in
spiral.
1. We need to print the nodes in BFS manner.
2. The nodes should be displayed in alternate direction; in one level
from left to right and in next level right to left.
Needless to mention, we need least time complex solution.
Thanks
Nayn
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If its just about display, you can traverse the tree in BFS manner
and
store the nodes in an array at their specific locations. At the end,
display
the nodes from the array.
Thanks,
--
Dinesh Bansal
The Law of Win says, Let's not do it your way or my way; let's do it
the
best way.
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