The problem never says that the tree is rooted. So LCA is not very relevant. The path between two nodes in any tree is unique. (Otherwise it has a cycle and is not a tree.) So all that's needed is to search for z starting at x (DFS or BFS will work fine). When you find the unique path, see if it contains y. This is O(n) where n is the number of nodes. The problem is more interesting if you are allowed to pre-process the tree one time in order to build a data structure to support many queries.
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