Deleting a Node from BST with Two Children

Start by clarifying the exact contract for Delete BST Node: input representation, mutation rules, duplicate behavior, and what should happen on invalid or empty input. The core idea is bst invariant preservation during two-child deletion. Work a tiny example aloud and name the invariant after each step, because that is the proof that the approach stays correct as the input grows.

A dependable implementation plan is to find target; choose inorder successor; replace target key; delete successor in right subtree; return changed root. The successor has no left child, which simplifies its deletion. Keep the code aligned with that invariant instead of adding special cases that silently change the contract. Test the smallest input, a normal case, duplicates or ties where relevant, an adversarial boundary, and a case that exercises the failure path. The expected complexity is O(h) time and O(h) recursion space. Explain both time and auxiliary memory, including recursion or helper structures; do not hide sorting, copying, or repeated scans. If the title omits critical constraints, say what you need to know before committing to a more specialized solution.