1. Let's define Bubble-Down time for a node of height $h$ as the number of comparisons or swaps needed until the node reaches a leaf. 2. Since at each step in Bubble-Down the node moves down one level, the time complexity for a node of height $h$ is proportional to $h$, i.e., $O(h)$. 3. Now, consider applying the Heapify operation on tree $T$. Nodes at level $h$ are all nodes with height $h$. 4. The number of nodes at height $h$ is approximately $\frac{n}{2^{h+1}}$ in a complete binary tree of size $n$, because level $h$ contains $2^{H-h}$ nodes and $H$ is the height of the tree. 5. Total time to Bubble-Down all nodes at level $h$ is then number of nodes times time per node, which is $$\frac{n}{2^{h+1}}\times O(h) = O\left(\frac{n h}{2^{h}}\right).$$ 6. Justification: Each node at height $h$ takes $O(h)$ time to Bubble-Down, and there are $\frac{n}{2^{h+1}}$ such nodes, resulting in total $O\left(\frac{n h}{2^{h}}\right)$ time complexity for all nodes at this level during Heapify.