In case of array representation of this binary tree where we can traverse
through all the leaf nodes and move to their parents, this problem becomes
quite easy.
for the example stated consider its array representation
arr[]={1,2,3,4,5,6,7} and take another array marked[7]={false} indicating
whether the current node has been printed
Now for every non leaf node, index i in (n/2) to (n-1)
print the leaf node and mark its marked[i]=true
change j= floor(i-1)/2
while marked[j] = false:
print arr[j]
marked[j]=true
change j=floor(j-1)/2
a C++ implementation for the above:-
int main(int argc, char **argv) {
int arr[] = { 1, 2, 3, 4, 5, 6, 7 }, N = 7;
bool marked[7];
for (int i = 0; i < N; i++)
marked[i] = false;
for (int i = N / 2; i < N; i++) {
printf("[%d]", arr[i]);
marked[i] = true;
for (int j = (i - 1) / 2; j>=0 && !marked[j]; j = (j - 1) / 2) {
printf("[%d]", arr[j]);
marked[j] = true;
}
}
}
Anurag Sharma
On Wed, Jun 9, 2010 at 9:31 PM, sharad kumar <sharad20073...@gmail.com>wrote:
>
>>>
>>>
>>> 1
>>> / \
>>> 2 3
>>> / \ / \
>>> 4 5 6 7
>>>
>>>
>>> print
>>>
>>> 4 2 1 5 6 3 7
>>>
>>>
> a set of vertical line 4m left u draw then on each line whatever no comes
> write it
>
>>
>>
>>
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>> S.Antony Vincent Pandian
>>
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