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 trick is a single post-order DFS. At each node, if it's null or its value equals p or q, return it. Otherwise recurse left and right. If both sides return something non-null, the current node is the split point, so it's the LCA. If only one side returns something, pass that up. That handles Example 2 naturally: hitting 5 returns immediately, and since the target 4 sits below it, 5 bubbles up as the answer. The early return on a match is what makes the self-descendant rule work for free. The common pitfall is recursion depth. With 100000 nodes, a skewed tree can blow the stack in some languages, so mention an iterative version using parent pointers if asked. Another pitfall is using BST logic on values. Don't. StealthCoder is the hedge if you freeze on the base case during the live OA, but this is about ten lines.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.