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A 4-ary tree,i.e. each node has either 0 or 4 children tree has 20 leaf nodes. Then the total number of nodes in the tree are ____.
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          /                      Root                               \                          |

        /                   /                  \                         \                         |

     N1                 N2                 N3                    N4                       N6 

/            \      /           \        /              \          /               \          /              \

1  2  3  4      5  6  7    8      9  10  11  12        13  14  15  16        17  18  19  20

from above scenario we can conclude that tree is not possible with given data...

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L=(K-1)*i+1

k=4

L=20

i=(20-1)/3

total node = internal + leaf nodes

N=i+l

N=20+7
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The above binary tree is not possible with above condition. However If the total number of leaf nodes were 19 instead of 20 then internal nodes would have been 6. And hence total number of nodes would have been

Internal + leaf node = 6+19=25
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20 leaf nodes arrangement for given constraint of 0 or 4 children ..

#    I:- no. Of internal nodes ...

##   L:- no. Of leaf node ...

###  n:- n- ary tree….

If u analyze some what you will get following formula:--»

(n-1) I +1 = L ...

But for Given question due to ur given constraint of 0 or 4 children ..

20 leaf nodes are not possible in this arrangment….

 

If 19 leaf nodes given then we have a solution for this :->> apply on above formula u will get 6 Internal nodes….

So total nodes in that case 19+6 = 25 nodes…

 

 

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