证明平衡二叉搜索树的高度为 log(n)

Question

The binary-search algorithm takes log(n) time, because of the fact that the height of the tree (with n nodes) would be log(n).

How would you prove this?

Answer 1

Now here I am not giving mathematical proof. Try to understand the problem using log to the base 2. Log ₂ is the normal meaning of log in computer science.

First, understand it is binary logarithm (log ₂ n) (logarithm to the base 2). For example,

• the binary logarithm of 1 is 0
• the binary logarithm of 2 is 1
• the binary logarithm of 3 is 1
• the binary logarithm of 4 is 2
• the binary logarithm of 5, 6, 7 is 2
• the binary logarithm of 8-15 is 3
• the binary logarithm of 16-31 is 4 and so on.

For each height the number of nodes in a fully balanced tree are

Height  Nodes  Log calculation
      0        1      log21 = 0
      1        3      log23 = 1
      2        7      log27 = 2
      3       15      log215 = 3

Consider a balanced tree with between 8 and 15 nodes (any number, let's say 10). It is always going to be height 3 because log ₂ of any number from 8 to 15 is 3.

In a balanced binary tree the size of the problem to be solved is halved with every iteration. Thus roughly log ₂ n iterations are needed to obtain a problem of size 1.

I hope this helps.

Answer 2

Let's assume at first that the tree is complete - it has 2^N leaf nodes. We try to prove that you need N recursive steps for a binary search.

With each recursion step you cut the number of candidate leaf nodes exactly by half (because our tree is complete). This means that after N halving operations there is exactly one candidate node left.

As each recursion step in our binary search algorithm corresponds to exactly one height level the height is exactly N.

Generalization to all balanced binary trees: If the tree has less nodes than 2^N we for sure don't need more halvings. We might need less or the same amount but never more.

Answer 3

Assuming that we have a complete tree to work with, we can say that at depth k, there are 2 ^k nodes. You can prove this using simple induction, based on the intuition that adding an extra level to the tree will increase the number of nodes in the entire tree by the number of nodes that were in the previous level times two.

The height k of the tree is log(N), where N is the number of nodes. This can be stated as

log ₂ (N) = k,

and it is equivalent to

N = 2 ^k

To understand this, here's an example:

16 = 2 ⁴ => log ₂ (16) = 4

The height of the tree and the number of nodes are related exponentially . Taking the log of the number of nodes just allows you to work backwards to find the height.

Answer 4

Just look up the rigorous proof in Knuth, Volume 3 - Searching and Sorting Algorithms ... He does it far more rigorously than anyone else I can think of.

http://en.wikipedia.org/wiki/The_Art_of_Computer_Programming

You can find it in any good Computer Science library and on the bookshelves of many (very) old geeks.

Answer 5

Why is the height of a balanced binary tree equal to ceil(log ₂ N) for N nodes?

w = width of base (maximum number of leaves)

h = height of tree (maximum number of edges from root to leaf)

Divide w by 2 (h times) to get to 1, which counts the single root node at top.

N = w + w/2 + ... + 1

N = 2 ^h + ... + 2 ¹ + 2 ⁰

= (1-2 ^h+1 ) / (1-2) = 2 ^h+1 -1

log ₂ (N+1) = h+1

Check: if N=1, h=0. If h=1, N=3.

This formula is for if the bottom level is full. N will not always be so great, but would still have the same height, h. So we must take the log's ceiling.