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The height of an AVL tree containing n nodes

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• Definitions and notations
  • Definitions:

    h = the height of an AVL tree n(h) = the minimum number of nodes in an AVL tree of height h

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• The value of n(1) and n(2)
  • Some facts on n(h):

    n(1) = 1 and n(2) = 2:

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• A recurrence relationship on n(h)
  • Consider the minimum number of nodes in an AVL tree of height h:

    Note:

    In order to put the minimum number of nodes in an AVL tree of height h, we must: put the minimum number of nodes in its 2 sub trees Because the maximum difference in height is 1 (one), the minimum sub trees are as follows: One of the sub tree has height h−1 (in order to make the height of the original tree = h) The other sub tree has height h−2 (to minimize the # nodes in the AVL tree) Both sub trees have the minimum number of nodes Therefore: Sub tree 1 has n(h−1) nodes Sub tree 2 has n(h−2) nodes

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  • Relationship between n(h), n(h−1) and n(h−2):

    n(h) = 1 + n(h−1) + n(h−2) ...... (1) n(1) = 1 n(2) = 2

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• Solving the recurrence relation
  • We use Mathematica to solve this recurrence equation:

    math In[0]:= RSolve[ {n[h]==1+n[h-1]+n[h-2], n[0]==1, n[1]==2}, n[h], h ] Out[0]= {{n[h] -> -1 + 2 Fibonacci[h] + LucasL[h]}} Which is equals to: / 1/2 \ / 1/2\h / 1/2 \ / 1/2\h |3 5 | | 5 | | 3 5 | | 5 | |------ + 1/2| |1/2 + ----| + |- ------ + 1/2| |1/2 - ----| - 1 \ 10 / \ 2 / \ 10 / \ 2 / 1/2 / 2 \h 1/2 / 2 \h 2 5 |- --------| 2 5 |- --------| | 1/2| | 1/2| \ 1 - 5 / \ 1 + 5 / - -------------------- + -------------------- 1/2 1/2 5 (1 - 5 ) 5 (1 + 5 ) approximately equal to ~= 2 × 1.6h

    Example:

    > -1 + 2 Fibonacci[5] + LucasL[5] 20 > 2*1.6^5 20.9715 > -1 + 2 Fibonacci[10] + LucasL[10] 232 > 2*1.6^10 219.902 > -1 + 2 Fibonacci[15] + LucasL[15] 2583 > 2*1.6^15 2305.84

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• The minimum number of nodes in an AVL tree of height h
  • From above, we found:

    n(h) = minimum number of nodes in an AVL tree of height h ~= 2 × 1.6h ....... (2)

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• Maximum height of an AVL tree with n nodes
  • Notation:

    m(h) = number of nodes in an AVL tree of height h

  • We have that:

    m(h) > min. # nodes in AVL tree of height h > n(h) > 2 × 1.6h Reverse the inequality: 2 × 1.6h < m(h) <==> 1.6h < m(h)/2 <==> log(1.6h) < log(m(h)/2) <==> h log(1.6) < log(m(h)) - 1 <==> 0.678 h < log(m(h)) - 1 <==> h < 1.475 log(m(h)) - 1.475 ...... (3) ==> h < 2 log(m(h)) + 2 (in text book)

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  • Therefore (acording to Goodrich):

    The height of an AVL tree that contains m entries is < 2 log(m) + 2

    (This was a rough estimate, I gave you a more accurate estimate above)

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• Examples
  • Some maximum heights:

    m = 100, h < 15.28 m = 1000, h < 21.93 m = 10000, h < 28.57

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• Running time of a search (lookup) operation in AVL tree
  • Running time of lookup:

    O(log(n))           (n = # entries)