CODEWITHSUNDEEP
algorithmtimetime-complexitybig-o
hstutea
hstutea

Reputation: 33

Big O for Height of Balanced Binary Tree

Perhaps a dumb question. In a balanced binary tree where n is the total number of nodes, I understand why the height is equal to log(n). What I don't understand is what people mean when they refer to the height as being O(log(n)). I've only seen Big O used in the context of algorithms, where if an algorithm runs in O(n) and if the input doubles then the running time doubles. But height isn't an algorithm. How does this apply to the height of a tree? What does it mean for the height to be O(log(n))?

Upvotes: 3

Views: 2159

Answers (3)

Patrick87
Patrick87

Reputation: 28332

To add to Berthur's excellent answer, Big-Oh notation is not specific to analysis of algorithms; it applies to any functions. In analysis f algorithms we care about the function T(n) which gives the (typically worst-case) runtime for input size n, and we want to know an upper bound (Big-Oh) on that function's rate of growth. Here, there is a function that gives the true height of a tree with whatever property, and we want an upper bound on that function's rate of growth. We could find upper bounds on arbitrary functions devoid of any context at all like f(n) = n!^(1/2^n) or whatever.

Upvotes: 1

Berthur
Berthur

Reputation: 4495

This is because a complete binary tree of n nodes does not have height log(n).

Consider a complete binary tree of height k. Such a tree has 2k leaf nodes. How many nodes does it have in total? If you look at each level, you will find that it has 1 + 2 + 4 + 8 + ... + 2k nodes, or 20 + 21 + 22 + 23 + ... 2k.

After some math, you will find that this series equals 2k+1 - 1.

So, if your tree has n nodes, what is its height? If you solve the equation n = 2k+1 - 1 with respect to k, you obtain k = log2(n+1) - 1.

This expression is slightly less nice than log2(n), and it is certainly not the same number. However, by the properties of big-O notation, log2(n+1) - 1 = O(log(n)).

In the source you are reading, emphasis is given on that the height grows as fast as log(n). They belong to the same complexity class. This information can be useful when designing algorithms, since you know that doubling the input will increase the tree height only by a constant. This property gives tree structures immense power. So even if the expression for the exact height of the tree is more complicated (and if your binary tree is not complete, it will look more complicated still), it is of logarithmic complexity with respect to n.

Upvotes: 2

Const
Const

Reputation: 9

I think they mean it takes O(log(n)) to traverse the tree

Upvotes: 0

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