One could alway store a node's depth at each node -- then you must search for u and v, creating a list of what nodes you found at each depth, and finally, simply compare the lists -- O(n) in the depth of u and v. Bob On 3 Dec 2007, at 08:40, apfelmus wrote:
Adrian Neumann wrote:
data Tree a = Leaf a | Node a [Tree a] But now the assignments require more than a simple top-down traversal. For example: given a tree t and two nodes u,v, find the first common ancestor.
Well, this problem doesn't make much sense in Haskell. How do you specify the subtrees u and v in the first place?
Regards, apfelmus
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