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Balanced Binary Search Trees
By iLoveIOI, history, 3 years ago, In English

How do you count the number of balanced binary search trees with N nodes? Balanced as in the left subtree size and the right subtree size differ by at most 1.

Thanks!

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3 years ago, # |
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If I am not wrong, then this number series might help — It is the number of possible balanced binary search tree with $$$N$$$ unique nodes

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    3 years ago, # ^ |
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    I don't think that is correct. 4 nodes has 4 trees N<=1e18 by the way

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      3 years ago, # ^ |
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      I think your mean was to deal with the problem of counting such balanced tree with exact $$$N$$$ nodes

      Lets take a subtree $$$G_0$$$ with $$$N$$$ nodes, including root. Since the subtree construction depend on left and right subsubtree, we have:

      • Excluding root, if $$$N - 1$$$ is even then each subsubtree will have $$$K = \frac{N - 1}{2}$$$ nodes
      • Excluding root, if $$$N - 1$$$ is odd then one subtree will have $$$K_1 = \frac{N - 1}{2}$$$ nodes and other will have $$$K_2 = N - K_1$$$

      Lets $$$f(x) = $$$ numbers of balanced tree with $$$N$$$ nodes, from above observation, we have

      • When $$$N - 1$$$ is even $$$f(N) = f(K)^2$$$
      • When $$$N - 1$$$ is odd $$$f(N) = 2 \times f(K_1) \times f(K_2)$$$
      Example
      f(0) -> f(100)
      Series
      Generator

      Since every value is the power of 2, we also have this generator for a funny graph

      Generator
      Graph - 256

      It is A110316 OEIS

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        3 years ago, # ^ |
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        Wow! Thanks! That helped a lot!

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          3 years ago, # ^ |
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          Then give me contribution UwU

          Just kidding, by the way, since there is $$$O(log_2(n))$$$ solution by matrix multiplication and $$$O(log_2^2(n))$$$ solutions by recursive-dp, do you think there is an $$$O(1)$$$ solution by combinatoric or something ?

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