Update 0261-graph-valid-tree.py

I'd propose to include DSU-based solution to this problem as it more efficient. If I'm not mistaken O(ElogV) would be time complexity and the approach saves some space as we don't recreate graph\tree into adjacency list prior dfs and loop over the edge list directly

python/0261-graph-valid-tree.py

@@ -29,3 +29,45 @@ def dfs(i, prev):
             return True
 
         return dfs(0, -1) and n == len(visit)
+
+
+
+    # alternative solution via DSU O(ElogV) time complexity and 
+    # save some space as we don't recreate graph\tree into adjacency list prior dfs and loop over the edge list directly
+    class Solution:
+    """
+    @param n: An integer
+    @param edges: a list of undirected edges
+    @return: true if it's a valid tree, or false
+    """
+    def __find(self, n: int) -> int:
+        while n != self.parents.get(n, n):
+            n = self.parents.get(n, n)
+        return n
+
+    def __connect(self, n: int, m: int) -> None:
+        pn = self.__find(n)
+        pm = self.__find(m)
+        if pn == pm:
+            return
+        if self.heights.get(pn, 1) > self.heights.get(pm, 1):
+            self.parents[pn] = pm
+        else:
+            self.parents[pm] = pn
+            self.heights[pm] = self.heights.get(pn, 1) + 1
+        self.components -= 1
+
+    def valid_tree(self, n: int, edges: List[List[int]]) -> bool:
+        # init here as not sure that ctor will be re-invoked in different tests
+        self.parents = {}
+        self.heights = {}
+        self.components = n
+
+        for e1, e2 in edges:
+            if self.__find(e1) == self.__find(e2):  # 'redundant' edge
+                return False
+            self.__connect(e1, e2)
+
+        return self.components == 1  # forest contains one tree
+
+