A question I've been struggling with...
Why the implementation of a 2-3 trees doesn’t allow nodes to have degree of 1?
I thought it might be related to the O(log(n)) it (as a member of the B tree family) wants to keep, if degree 1 was allowed we could get a tree like that:
1
\
2
\
3
\
4
\
5
for example and then some operations will take O(n) instead of O(log(n))
but i don't see where in this answer i referred to a 2-3 tree and why it can't allow degree 1... :-/
thanks! ;-)
You already have the right answer, but maybe you want to say it like this:
B-tree variants maintain all leaves at the same depth (the tree
height), and operations generally take time proportional to that
height.
Since internal nodes must have at least 2 children, each
level contains at least twice as many nodes as the parent level, and
the height of the tree is O(log N).
If internal nodes were allowed to contain
less than 2 children, the height could be more than O(log N) and
operations would take longer than logarithmic time.
Related
Does a skewed binary tree take more space than, say, a perfect binary tree ?
I was solving the question #654 - Maximum Binary Tree on Leetcode, where given an array you gotta make a binary tree such that, the root is the maximum number in the array and the right and left sub-tree are made on the same principle by the sub-array on the right and left of the max number, and there its concluded that in average and best case(perfect binary tree) the space taken would be O(log(n)), and worst case(skewed binary tree) would be O(n).
For example, given nums = [1,3,2,7,4,6,5],
the tree would be as such,
7
/ \
3 6
/ \ / \
1 2 4 5
and if given nums = [7,6,5,4,3,2,1],
the tree would be as such,
7
\
6
/ \
5
/ \
4
/ \
3
/ \
2
/ \
1
According to my understanding they both should take O(n) space, since they both have n nodes. So i don't understand how they come to that conclusion.
Thanks in advance.
https://leetcode.com/problems/maximum-binary-tree/solution/
Under "Space complexity," it says:
Space complexity : O(n). The size of the set can grow upto n in the worst case. In the average case, the size will be nlogn for n elements in nums, giving an average case complexity of O(logn).
It's poorly worded, but it is correct. It's talking about the amount of memory required during construction of the tree, not the amount of memory that the tree itself occupies. As you correctly pointed out, the tree itself will occupy O(n) space, regardless if it's balanced or degenerate.
Consider the array [1,2,3,4,5,6,7]. You want the root to be the highest number, and the left to be everything that's to the left of the highest number in the array. Since the array is in ascending order, what happens is that you extract the 7 for the root, and then make a recursive call to construct the left subtree. Then you extract the 6 and make another recursive call to construct that node's left subtree. You continue making recursive calls until you place the 1. In all, you have six nested recursive calls: O(n).
Now look what happens if your initial array is [1,3,2,7,5,6,4]. You first place the 7, then make a recursive call with the subarray [1,3,2]. Then you place the 3 and make a recursive call to place the 1. Your tree is:
7
3
1
At this point, your call depth is 2. You return and place the 2. Then return from the two recursive calls. The tree is now:
7
3
1 2
Constructing the right subtree also requires a call depth of 2. At no point is the call depth more than two. That's O(log n).
It turns out that the call stack depth is the same as the tree's height. The height of a perfect tree is O(log n), and the height of a degenerate tree is O(n).
Consider a binary tree and some traverse criterion that defines an ordering of the tree's elements.
Does it exists some particular traverse criterion that would allow a begin_insert operation, i.e. the operation of adding a new element that would be at position 1 according to the ordering induced by the traverse criterion, with O(log(N)) cost?
I don't have any strict requirement, like the tree guaranteed to be balanced.
EDIT:
But I cannot accept lack of balance if that allows degeneration to O(N) in worst case scenarios.
EXAMPLE:
Let's try to see if in-order traversal would work.
Consider the BT (not a binary search tree)
6
/ \
13 5
/ \ /
2 8 9
In-order traversal gives 2-13-8-6-9-5
Perform begin_insert(7) in such a way that in-order traversal gives 7-2-13-8-6-9-5:
6
/ \
13 5
/ \ /
2 8 9
/
7
Now, I think this is not a legitimate O(log(N)) strategy, because if I keep adding values in this way the cost degenerates into O(N) as the tree becomes increasingly unbalanced
6
/ \
13 5
/ \ /
2 8 9
/
7
/
*
/
*
/
This strategy would work if I rebalance the tree by preserving ordering:
8
/ \
2 9
/ \ / \
7 13 6 5
but this costs at least O(N).
According to this example my conclusion would be that in-order traversal does not solve the problem, but since I received feedback that it should work maybe I am missing something?
Inserting, deleting and finding in a binary tree all rely on the same search algorithm to find the right position to do the operation. The complexity of this O(max height of the tree). The reason is that to find the right location you start at the root node and compare keys to decide if you should go into the left subtree or the right subtree and you do this until you find the right location. The worst case is when you have to travel down the longest chain which is also the definition for height of the tree.
If you don't have any constraints and allow any tree then this is going to be O(N) since you allow a tree with only left children (for example).
If you want to get better guarantees you must use algorithms that promise that the height of the tree has a lower bound. For example AVL guarantees that your tree is balanced so the max height is always log N and all the operations above run in O(log N). Red-black trees don't guarantee log N but promise that the tree is not going to be too unbalanced (min height * 2 >= max height) so it keeps O(log N) complexity (see page for details).
Depending on your usage patterns you might be able to find more specialized data structures that give even better complexity (see Fibonacci heap).
I can see how, when looking up a value in a BST we leave half the tree everytime we compare a node with the value we are looking for.
However I fail to see why the time complexity is O(log(n)). So, my question is:
If we have a tree of N elements, why the time complexity of looking up the tree and check if a particular value exists is O(log(n)), how do we get that?
Your question seems to be well answered here but to summarise in relation to your specific question it might be better to think of it in reverse; "what happens to the BST solution time as the number of nodes goes up"?
Essentially, in a BST every time you double the number of nodes you only increase the number of steps to solution by one. To extend this, four times the nodes gives two extra steps. Eight times the nodes gives three extra steps. Sixteen times the nodes gives four extra steps. And so on.
The base 2 log of the first number in these pairs is the second number in these pairs. It's base 2 log because this is a binary search (you halve the problem space each step).
For me the easiest way was to look at a graph of log2(n), where n is the number of nodes in the binary tree. As a table this looks like:
log2(n) = d
log2(1) = 0
log2(2) = 1
log2(4) = 2
log2(8) = 3
log2(16)= 4
log2(32)= 5
log2(64)= 6
and then I draw a little binary tree, this one goes from depth d=0 to d=3:
d=0 O
/ \
d=1 R B
/\ /\
d=2 R B R B
/\ /\ /\ /\
d=3 R B RB RB R B
So as the number of nodes, n, in the tree effectively doubles (e.g. n increases by 8 as it goes from 7 to 15 (which is almost a doubling) when the depth d goes from d=2 to d=3, increasing by 1.) So the additional amount of processing required (or time required) increases by only 1 additional computation (or iteration), because the amount of processing is related to d.
We can see that we go down only 1 additional level of depth d, from d=2 to d=3, to find the node we want out of all the nodes n, after doubling the number of nodes. This is true because we've now searched the whole tree, well, the half of it that we needed to search to find the node we wanted.
We can write this as d = log2(n), where d tells us how much computation (how many iterations) we need to do (on average) to reach any node in the tree, when there are n nodes in the tree.
This can be shown mathematically very easily.
Before I present that, let me clarify something. The complexity of lookup or find in a balanced binary search tree is O(log(n)). For a binary search tree in general, it is O(n). I'll show both below.
In a balanced binary search tree, in the worst case, the value I am looking for is in the leaf of the tree. I'll basically traverse from root to the leaf, by looking at each layer of the tree only once -due to the ordered structure of BSTs. Therefore, the number of searches I need to do is number of layers of the tree. Hence the problem boils down to finding a closed-form expression for the number of layers of a tree with n nodes.
This is where we'll do a simple induction. A tree with only 1 layer has only 1 node. A tree of 2 layers has 1+2 nodes. 3 layers 1+2+4 nodes etc. The pattern is clear: A tree with k layers has exactly
n=2^0+2^1+...+2^{k-1}
nodes. This is a geometric series, which implies
n=2^k-1,
equivalently:
k = log(n+1)
We know that big-oh is interested in large values of n, hence constants are irrelevant. Hence the O(log(n)) complexity.
I'll give another -much shorter- way to show the same result. Since while looking for a value we constantly split the tree into two halves, and we have to do this k times, where k is number of layers, the following is true:
(n+1)/2^k = 1,
which implies the exact same result. You have to convince yourself about where that +1 in n+1 is coming from, but it is okay even if you don't pay attention to it, since we are talking about large values of n.
Now let's discuss the general binary search tree. In the worst case, it is perfectly unbalanced, meaning all of its nodes has only one child (and it becomes a linked list) See e.g. https://www.cs.auckland.ac.nz/~jmor159/PLDS210/niemann/s_fig33.gif
In this case, to find the value in the leaf, I need to iterate on all nodes, hence O(n).
A final note is that these complexities hold true for not only find, but also insert and delete operations.
(I'll edit my equations with better-looking Latex math styling when I reach 10 rep points. SO won't let me right now.)
Whenever you see a runtime that has an O(log n) factor in it, there's a very good chance that you're looking at something of the form "keep dividing the size of some object by a constant." So probably the best way to think about this question is - as you're doing lookups in a binary search tree, what exactly is it that's getting cut down by a constant factor, and what exactly is that constant?
For starters, let's imagine that you have a perfectly balanced binary tree, something that looks like this:
*
/ \
* *
/ \ / \
* * * *
/ \ / \ / \ / \
* * * * * * * *
At each point in doing the search, you look at the current node. If it's the one you're looking for, great! You're totally done. On the other hand, if it isn't, then you either descend into the left subtree or the right subtree and then repeat this process.
If you walk into one of the two subtrees, you're essentially saying "I don't care at all about what's in that other subtree." You're throwing all the nodes in it away. And how many nodes are in there? Well, with a quick visual inspection - ideally one followed up with some nice math - you'll see that you're tossing out about half the nodes in the tree.
This means that at each step in a lookup, you either (1) find the node that you're looking for, or (2) toss out half the nodes in the tree. Since you're doing a constant amount of work at each step, you're looking at the hallmark behavior of O(log n) behavior - the work drops by a constant factor at each step, and so it can only do so logarithmically many times.
Now of course, not all trees look like this. AVL trees have the fun property that each time you descend down into a subtree, you throw away roughly a golden ratio fraction of the total nodes. This therefore guarantees you can only take logarithmically many steps before you run out of nodes - hence the O(log n) height. In a red/black tree, each step throws away (roughly) a quarter of the total nodes, and since you're shrinking by a constant factor you again get the O(log n) lookup time you'd like. The very fun scapegoat tree has a tuneable parameter that's used to determine how tightly balanced it is, but again you can show that every step you take throws away some constant factor based on this tuneable parameter, giving O(log n) lookups.
However, this analysis breaks down for imbalanced trees. If you have a purely degenerate tree - one where every node has exactly one child - then every step down the tree that you take only tosses away a single node, not a constant fraction. That means that the lookup time gets up to O(n) in the worst case, since the number of times you can subtract a constant from n is O(n).
If we have a tree of N elements, why the time complexity of looking up
the tree and check if a particular value exists is O(log(n)), how do
we get that?
That's not true. By default, a lookup in a Binary Search Tree is not O(log(n)), where n is a number of nodes. In the worst case, it can become O(n). For instance, if we insert values of the following sequence n, n - 1, ..., 1 (in the same order), then the tree will be represented as below:
n
/
n - 1
/
n - 2
/
...
1
A lookup for a node with value 1 has O(n) time complexity.
To make a lookup more efficient, the tree must be balanced so that its maximum height is proportional to log(n). In such case, the time complexity of lookup is O(log(n)) because finding any leaf is bounded by log(n) operations.
But again, not every Binary Search Tree is a Balanced Binary Search Tree. You must balance it to guarantee the O(log(n)) time complexity.
assume I have a complete binary tree up-to a certain depth d. What would the time complexity be to traverse (pre-order traversal) this tree.
I am confused because I know that the amount of nodes in the tree is 2^d, so therefore the time complexity would be BigO(2^d) ? because the tree is growing exponentially.
But, upon research on the internet, Everyone states that's traversal is BigO(n) where n is the number of elements (which would be 2^d in this case), not BigO(2^d), what am I missing?
thanks
n is defined as the number of nodes.
2^d is only the number of nodes when every possible node at that depth is full
ie.
o
/ \
o o
/ \
o o
only has 5 nodes when 2^d is 8
A complete binary tree has every node filled except for last row and all of the nodes are filled to the left. You can find the definition on wikipedia
http://en.wikipedia.org/wiki/Binary_tree#Types_of_binary_trees
Even if you can express the time complexity as O(2^d), that's pretty useless as it's not something that you can use to compare it to the time complexity of any other collection.
Expressing the time complexity as O(n) is on the other hand very useful. It tells you exactly how the collection reacts when you increase the number of items, without having to know exactly how the collection is implement, and you can compare it to the time complexity of other collections.
Why is it important that a binary tree be balanced
Imagine a tree that looks like this:
A
\
B
\
C
\
D
\
E
This is a valid binary tree, but now most operations are O(n) instead of O(lg n).
The balance of a binary tree is governed by the property called skewness. If a tree is more skewed, then the time complexity to access an element of a the binary tree increases. Say a tree
1
/ \
2 3
\ \
7 4
\
5
\
6
The above is also a binary tree, but right skewed. It has 7 elements, so an ideal binary tree require O(log 7) = 3 lookups. But you need to go one more level deep = 4 lookups in worst case. So the skewness here is a constant 1. But consider if the tree has thousands of nodes. The skewness will be even more considerable in that case. So it is important to keep the binary tree balanced.
But again the skewness is the topic of debate as the probablity analysis of a random binary tree shows that the average depth of a random binary tree with n elements is 4.3 log n . So it is really the matter of balancing vs the skewness.
One more interesting thing, computer scientists have even found an advantage in the skewness and proposed a skewed datastructure called skew heap
To ensure log(n) search time, you need to divide the total number of down level nodes by 2 at each branch. For example, if you have a linear tree, never branching from root to the leaf node, then the search time will be linear as in a linked list.
An extremely unbalanced tree, for example a tree where all nodes are linked to the left, means you still search through every single node before finding the last one, which is not the point of a tree at all and has no benefit over a linked list. Balancing the tree makes for better search times O(log(n)) as opposed to O(n).
As we know that most of the operations on Binary Search Trees proportional to height of the Tree, So it is desirable to keep height small. It ensure that search time strict to O(log(n)) of complexity.
Rather than that most of the Tree Balancing Techniques available applies more to
trees which are perfectly full or close to being perfectly balanced.
At the end of the end you need the simplicity over your tree and go for best binary trees like red-black tree or avl