#### Eulerian Path and Circuit for Undirected Graph

W
```# Eulerian Path is a path in graph that visits every edge exactly once.
# Eulerian Circuit is an Eulerian Path which starts and ends on the same
# vertex.
# time complexity is O(V+E)
# space complexity is O(VE)

# using dfs for finding eulerian path traversal
def dfs(u, graph, visited_edge, path=None):
path = (path or []) + [u]
for v in graph[u]:
if visited_edge[u][v] is False:
visited_edge[u][v], visited_edge[v][u] = True, True
path = dfs(v, graph, visited_edge, path)
return path

# for checking in graph has euler path or circuit
def check_circuit_or_path(graph, max_node):
odd_degree_nodes = 0
odd_node = -1
for i in range(max_node):
if i not in graph.keys():
continue
if len(graph[i]) % 2 == 1:
odd_degree_nodes += 1
odd_node = i
if odd_degree_nodes == 0:
return 1, odd_node
if odd_degree_nodes == 2:
return 2, odd_node
return 3, odd_node

def check_euler(graph, max_node):
visited_edge = [[False for _ in range(max_node + 1)] for _ in range(max_node + 1)]
check, odd_node = check_circuit_or_path(graph, max_node)
if check == 3:
print("graph is not Eulerian")
print("no path")
return
start_node = 1
if check == 2:
start_node = odd_node
print("graph has a Euler path")
if check == 1:
print("graph has a Euler cycle")
path = dfs(start_node, graph, visited_edge)
print(path)

def main():
g1 = {1: [2, 3, 4], 2: [1, 3], 3: [1, 2], 4: [1, 5], 5: [4]}
g2 = {1: [2, 3, 4, 5], 2: [1, 3], 3: [1, 2], 4: [1, 5], 5: [1, 4]}
g3 = {1: [2, 3, 4], 2: [1, 3, 4], 3: [1, 2], 4: [1, 2, 5], 5: [4]}
g4 = {1: [2, 3], 2: [1, 3], 3: [1, 2]}
g5 = {
1: [],
2: []
# all degree is zero
}
max_node = 10
check_euler(g1, max_node)
check_euler(g2, max_node)
check_euler(g3, max_node)
check_euler(g4, max_node)
check_euler(g5, max_node)

if __name__ == "__main__":
main()
```