A Question of Some Depth
Let Tbe a k-ary tree, i.e. each internal node of Thas at most kchildren.
Suppose that the depth of Tis d(the maximum depth of any node of Tis
d). What is the maximum number of external nodes (leaves) that Tcan have?
What is the maximum total number of nodes that Tcan have?
Solution: The number of external nodes and the total number of nodes will
clearly be maximized when every internal node has the maximum number of
children.
For a k-ary tree, the root node then has kchildren. Each of these children
will have kchildren of their own, so there will be k2children at depth 2 in T.
Similarly, each of these k2nodes will have kchildren, so there are k2·k=k3
vertices at depth 3, and so forth. The leaves of the tree will have depth d.
Therefore, there will be kdnodes (leaves) at depth d. So the maximum number
of leaves for a k-ary tree with depth dis kd.
Then the maximum total number of nodes is then
1 + k+k2+k3+···+kd=kd+1 −1
k−1.
Note: For your information, here we show that (assuming k6= 1)
1 + k+k2+k3+···+kd=kd+1 −1
k−1.
(I don’t necessarily expect you to remember this formula, but more importantly,
if you wanted to reproduce it, you should remember the idea behind you could
get it.)
Let
S=kd+kd−1+· · · +k2+k+ 1.
Then
kS =kd+1 +kd+· · · +k3+k2+k .
Therefore
kS −S= (kd+1 +kd+· · · +k2+k)−(kd+kd−1+· · · +k+ 1)
and hence
(k−1)S=kd+1 −1.
So (dividing by (k−1) which is allowed since k6= 1) we get that
S=kd+1 −1
k−1.
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