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To combine more and less significant digital comparators, compute $E_H=\neg (L_H\vee G_H)$ and use the displayed formulas. Unequal upper blocks decide the order, because their difference outweighs every possible lower-block difference; equal upper blocks delegate the decision to the lower block. This is an associative operation on the three comparison outcomes. A balanced binary tree of these combinations has $O(n)$ circuit size and $O(\log n)$ circuit depth for $n$-bit inputs.

Yes, provided balancing guarantees logarithmic height. In a suitably maintained balanced binary tree of height $O(\log (n+1))$, insertion searches one root-to-leaf path and deletion of the minimum follows the leftmost path. Reconnecting the tree and performing the required rebalancing also costs at most a constant amount per level for the standard logarithmic-time schemes. Thus both operations take $O(\log (n+1))$ time, conventionally written $O(\log n)$ for nontrivial sizes. Merely calling a tree approximately balanced would not prove that height bound; it is the balancing guarantee that matters. Without it, a binary search tree can have linear height.

Apply comparator composition recursively in a balanced binary tree, with one-bit digital comparators at its leaves. For a power of two $n$, the circuit size and circuit depth satisfyThere are $n$ leaves and $n-1$ internal nodes, so the circuit size is $O(n)$ and the combinational delay is $O(\log n)$. Here each fixed-fan-in logic gate has bounded delay. For arbitrary $n$, pad both inputs with leading zero bits to the next power of two, which is less than $2n$; this preserves their unsigned interpretation and both bounds.

Choose the middle element as the root:The input is a finite, strictly ascending list, as required for distinct elements of a set. The two child lengths differ by at most one. Applying the same argument recursively gives a balanced binary tree of logarithmic height. All values on each side retain their original order, proving the binary search tree property. With linked lists, the repeated splitting gives $O(n\log n)$ time; it does not repeatedly insert values into a skewed tree.

Expand a seed by a function returning either no node or a left seed, root label and right seed. Recursively expanding the two seeds produces a binary tree. A finite ascending list split about its midpoint supplies an unfold into a balanced binary tree that is a binary search tree.