Lowest Common Ancestor in a Binary Tree

Given a binary tree and two distinct nodes in it, return the value of their lowest common ancestor. A node counts as a descendant of itself.
For this exercise, the runner supplies the target nodes by their unique integer values p and q, and returns the ancestor value instead of a node reference. Both targets exist in the tree. The tree is an ordinary binary tree, not necessarily a binary search tree.

Function
lowestCommonAncestorValue(root: TreeNode, p: int, q: int) → int

Examples
Example 1
root = [3,5,1,6,2,0,8,null,null,7,4]
p = 5
q = 1
return = 3
The first common ancestor is the root.
Example 2
root = [3,5,1,6,2,0,8,null,null,7,4]
p = 5
q = 4
return = 5
A node can be its own descendant, so 5 is the common ancestor.
Example 3
root = [1,2]
p = 1
q = 2
return = 1
The root is one of the requested nodes.

Constraints
The tree contains 2 through 100000 nodes.
-1000000000 <= node.val <= 1000000000; all node values are unique.
p != q and both target values occur in the tree.

The pattern is a post-order DFS. At each node, if it's null, or its value equals p or q, return it immediately. Otherwise recurse left and right. If both sides return non-null, the current node is the split point, so it's the answer. If only one side returns something, pass that up. That early return on a match is what covers the self-descendant case: when p sits above q, you return p and never need to look deeper. The common pitfall is searching for both nodes separately or assuming BST ordering. Another one is recursion depth. With up to 100000 nodes, a skewed tree can overflow the stack in some languages, so mention an iterative approach with parent pointers if asked. Time is O(n), space is O(h). If the recursion logic slips mid-assessment, StealthCoder is there as the hedge.

StealthCoder is the hedge for the one pattern you didn't drill. It runs invisibly during the screen share.