I have multiple binary trees stored as an array. In each slot is either nil (or null; pick your language) or a fixed tuple storing two numbers: the indices of the two "children". No node will have only one child -- it's either none or two.
Think of each slot as a binary node that only stores pointers to its children, and no inherent value.
Take this system of binary trees:
0 1
/ \ / \
2 3 4 5
/ \ / \
6 7 8 9
/ \
10 11
The associated array would be:
0 1 2 3 4 5 6 7 8 9 10 11
[ [2,3] , [4,5] , [6,7] , nil , nil , [8,9] , nil , [10,11] , nil , nil , nil , nil ]
I've already written simple functions to find direct parents of nodes (simply by searching from the front until there is a node that contains the child)
Furthermore, let us say that at relevant times, both all trees are anywhere between a few to a few thousand levels deep.
I'd like to find a function
P(m,n)
to find the lowest common ancestor of m and n -- to put more formally, the LCA is defined as the "lowest", or deepest node in which have m and n as descendants (children, or children of children, etc.). If there is none, a nil would be a valid return.
Some examples, given our given tree:
P( 6,11) # => 2
P( 3,10) # => 0
P( 8, 6) # => nil
P( 2,11) # => 2
The main method I've been able to find is one that uses an Euler trace, which turns the given tree, with a node A to be the invisible parent of 0 and 1 with a depth of -1, into:
A-0-2-6-2-7-10-7-11-7-2-0-3-0-A-1-4-1-5-8-5-9-5-1-A
And from that, simply find the node between your given m and n that has the lowest number; For example, to find P(6,11), look for a 6 and an 11 on the trace. The number between them that is the lowest is 2, and that's your answer. If A is in between them, return nil.
-- Calculating P(6,11) --
A-0-2-6-2-7-10-7-11-7-2-0-3-0-A-1-4-1-5-8-5-9-5-1-A
^ ^ ^
| | |
m lowest n
Unfortunately, I do believe that finding the Euler trace of a tree that can be several thousands of levels deep is a bit machine-taxing...and because my tree is constantly being changed throughout the course of the programming, every time I wanted to find the LCA, I'd have to re-calculate the Euler trace and hold it in memory every time.
Is there a more memory efficient way, given the framework I'm using? One that maybe iterates upwards? One way I could think of would be the "count" the generation/depth of both nodes, and climb the lowest node until it matched the depth of the highest, and increment both until they find someone similar.
But that'd involve climbing up from level, say, 3025, back to 0, twice, to count the generation, and using a terribly inefficient climbing-up algorithm in the first place, and then re-climbing back up.
Are there any other better ways?
