# How is the number of chains formed in heavy light decomposition of a tree=log(n)?

I am not able to understand the proof given in http://wcipeg.com/wiki/Heavy-light_decomposition

My understanding is when the tree is heavily unbalanced the number of chains that can be formed is in the order of n/2.

Please correct me if I am wrong.

for heavily unbalanced i guess it will be lesser numbers of chains.
take example fully right skewed tree.
parent -> child
start -> 1 (root node)
1 -> 2
2 -> 3
3 -> 4
4 -> 5

Now when you make chains in this tree, you are going to have only one segment. that is 1 to 5.

similarly take example of balanced tree

start - > 1
1 -> 2
1 -> 3
2 -> 4
2 -> 5
3 -> 6
3 -> 7

in this graph your chains are going to look like :

You can see number of chains now are 4.

@sirearsh letâ€™s take this example below,

1->2
1->3
3->4
3->5
5->6
5->7
7->8
7->9
.
.
and so on,
according to my understanding the odd nodes form a single long chain and all the even nodes form n/2 chains of length 1. total (n/2 + 1) chains.

Am I correct ?