# Worst case bisection of binary tree

My algorithm book states that any n-vertex binary tree T can be partitioned by just removing a single edge into two disconnected trees A and B where neither of them has more than 3/4 of the vertices.

It sounds like it should be simple to create such a tree, but I can't imagine one, my bisections are always better balanced. Can somebody show me a tree where the vertex distribution of 3/4 to 1/4?

This is from "Introduction to Algorithms" by Thomas Cormen, 3rd edition, MIT Press. Appendix B, Problems B-3.

Consider a simple binary tree $T$ with only 4 nodes: The root of $T$ is $A$. $A$ has a left child $B$ which has two children $C$ and $D$.
Nobody said that $3/4$ was the optimal number. In fact, it seems like $2/3$ might be achievable. Starting at the root, consider a walk which always chooses the child with the larger subtree. Consider the first time that you reach a node whose subtree contains less than $2/3$ of the nodes. Its parent's subtree contains more than $2/3$ of the nodes, so the subtree rooted at its larger child contains at least $1/3$ of the nodes. So we have found a node whose subtree contains between $1/3$ and $2/3$ of the nodes.
• I think the little flaw of your argument lies in the "rounding" of rational numbers to integers. (Consider a simple binary tree $T$ with only 4 nodes: The root of $T$ is $A$. $A$ has a left child $B$ which has two children $C$ and $D$.) – hengxin May 16 '15 at 7:40