## Problem Given a binary tree and two target values, return the value of their lowest common ancestor only if both targets occur in the tree. Return `null` if either target is absent. ### Function Contract Implement `lowestCommonAncestorIfPresent(root, p, q)`. - Node values are unique. - If `p == q`, return that value only if it occurs in the tree. - An ancestor may be one of the target nodes. ### Constraints & Assumptions - The tree has at most `100,000` nodes. - Node and target values are signed 32-bit integers. - The tree may be empty or highly skewed. ### Clarifying Questions to Ask - Are both targets guaranteed to exist? No; this is the central edge case. - Are values unique? Yes, so targets may be identified by value. - What if the targets are equal? Presence of that one value is sufficient. - Is recursion-depth failure relevant? Yes for a depth-`100,000` tree; an iterative solution is acceptable. ```hint Return evidence with the candidate A subtree result can carry a candidate ancestor plus flags or a count showing which targets were actually found. ``` ```hint Do not trust the standard shortcut alone Returning a found target immediately can produce that target when the other target is missing; validate presence before returning the candidate. ``` ### Example ```text tree = [3,5,1,6,2,0,8,null,null,7,4] p = 5 q = 4 output = 5 ``` For the same tree with `q = 42`, return `null`. ### Evaluation Focus - Distinguishes a valid ancestor from a partial match. - Handles equal targets, an ancestor target, missing nodes, and an empty tree. - Visits each node at most a constant number of times. - Explains recursive stack risk and an iterative alternative. ### Extensions to Discuss 1. How would parent pointers reduce the problem to intersecting ancestor chains? 2. How would you answer many LCA queries on one static tree? 3. What changes when values are not unique and targets are node identities?