Thus, the computation for the leaves are identical to
that in the binary tree.
If , then
We now consider non-leaf nodes . Let be the number of children of , let be its children from left to right, and let the subtrees rooted at the children of be respectively. Denote by the subtree of induced by the vertex set of , for . Denote by the total number of leaves in .
Define , where or , , , and , as follows. First let . If , is the minimal cost of a pebbling placement of the subtree , where we use at most pebbles in , and no pebble is placed on . (When no feasible pebbling placement exists with this constraint we have .) If , is the same as above except is placed with a pebble out of pebbles.
This definition is generalized for . For , we consider in place of and for we stipulate that one additional pebble is placed at the tip of the external path from of distance from . As before this additional pebble is not counted in .
We then define
Again we will compute for all , inductively for according to the height of the subtree , starting with leaves . The base case having already been taken care of, we assume and .
First we consider the left most subtree with , i.e., we compute for .
If and , then
Similarly for and , then
Now we consider . Again if ,
Similarly for and ,
We proceed to the case of . This time we inductively assume that we have already computed not only all with , but also the relevant quantities for .
Thus, for and ,
For and ,
Finally we consider the case and
.
For and , we have
And for and , we have
We have completed the description of the algorithm.
The final answer is ,
where is the root of and is the number of pebbles.
Again, no need to compute for any value ,
if is the total number of pebbles given.
The complexity of the algorithm can be easily estimated as before. For leaves, the algorithm spends time per leaf. Thus the total work spent on leaves is at most . For any non-leaf , suppose the degree of is , then the computation work spent for is . Thus the total amount of work spent for non-leaves is . Hence the total running time is at most , which is again only with .