Question 44

A complete binary min-heap stores the integers $[1, 1023]$ with the root at depth 0.

In a min-heap, the key of every parent is strictly less than the keys of its children. Along any simple path from the root to a node at depth $d$, the values must strictly increase:

If node 9 is located at depth $d$, there must be exactly $d$ ancestor nodes on the path from the root to node 9, and every ancestor must have a value strictly less than 9.

The set of available integers in the heap strictly less than 9 is $\{1, 2, 3, 4, 5, 6, 7, 8\}$, which contains exactly 8 values. Thus, the path can have at most 8 ancestors:

This places 9 at depth 8.

Since $N = 1023 = 2^{10} - 1$, the tree is completely full up to depth 9 (containing $2^9 = 512$ leaves at depth 9 and $2^8 = 256$ nodes at depth 8). Therefore, depth 8 is well within the tree bounds and can be realized by placing the remaining 1014 larger elements in other branches.

Thus, the maximum depth at which integer 9 can appear is 8.