# Exercise 6.2.6

Show that the worst-case running time of MAX-HEAPIFY on a heap of size $n$ is $\Omega(\lg{n})$. (Hint: For a heap with $n$ nodes, give node values that cause MAX-HEAPIFY to be called recursively at every node on a simple path from the root down to a leaf.)

This is another obvious one.

Let's take the leftmost path in the given heap. If we put the largest elements in the heap on that path and the smallest one at the root, MAX-HEAPIFY will need to be invoked once for each level in the heap in order to push the minimum element to the leftmost leaf. Since the heap height (and thus the leftmost path's length) is $\lfloor \lg{n} \rfloor$ (exercise 6.1-2), the worst-case running time of the procedure is $\Omega(\lg{n})$.