graph-valid-tree.py

# Time:  O(|V| + |E|)
# Space: O(|V| + |E|)

# BFS solution. Same complexity but faster version.
class Solution:
    # @param {integer} n
    # @param {integer[][]} edges
    # @return {boolean}
    def validTree(self, n, edges):
        if len(edges) != n - 1:  # Check number of edges.
            return False
        elif n == 1:
            return True

        visited_from, neighbors = 0, 1

        # A structure to track each node's [visited_from, neighbors]
        nodes = collections.defaultdict(lambda: [-1, []])
        for edge in edges:
            nodes[edge[0]][neighbors].append(edge[1])
            nodes[edge[1]][neighbors].append(edge[0])
        
        if len(nodes) != n:  # Check number of nodes.
            return False

        # BFS to check whether the graph is valid tree.
        visited = {}
        q = collections.deque()
        q.append(0)
        while q:
            i = q.popleft()
            visited[i] = True
            for node in nodes[i][neighbors]:
                if node != nodes[i][visited_from]:
                    if node in visited:
                        return False
                    else:
                        visited[node] = True
                        nodes[node][visited_from] = i
                        q.append(node)
        return len(visited) == n


# Time:  O(|V| + |E|)
# Space: O(|V| + |E|)
# BFS solution.
class Solution2:
    # @param {integer} n
    # @param {integer[][]} edges
    # @return {boolean}
    def validTree(self, n, edges):
        visited_from, neighbors = 0, 1

        # A structure to track each node's [visited_from, neighbors]
        nodes = collections.defaultdict(lambda: [-1, []])
        for edge in edges:
            nodes[edge[0]][neighbors].append(edge[1])
            nodes[edge[1]][neighbors].append(edge[0])

        # BFS to check whether the graph is valid tree.
        visited = {}
        q = collections.deque()
        q.append(0)
        while q:
            i = q.popleft()
            visited[i] = True
            for node in nodes[i][neighbors]:
                if node != nodes[i][visited_from]:
                    if node in visited:
                        return False
                    else:
                        visited[node] = True
                        nodes[node][visited_from] = i
                        q.append(node)
        return len(visited) == n