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# How to implement depth-first search in Python Educative Answers Team

Depth-first search (DFS), is an algorithm for tree traversal on graph or tree data structures. It can be implemented easily using recursion and data structures like dictionaries and sets.

## The Algorithm

1. Pick any node. If it is unvisited, mark it as visited and recur on all its adjacent nodes.
2. Repeat until all the nodes are visited, or the node to be searched is found.

## Implementation

Consider this graph, implemented in the code below: # Using a Python dictionary to act as an adjacency list
graph = {
'A' : ['B','C'],
'B' : ['D', 'E'],
'C' : ['F'],
'D' : [],
'E' : ['F'],
'F' : []
}

visited = set() # Set to keep track of visited nodes.

def dfs(visited, graph, node):
if node not in visited:
print (node)
for neighbour in graph[node]:
dfs(visited, graph, neighbour)

# Driver Code
dfs(visited, graph, 'A')

## Explanation

• Lines 2-9: The illustrated graph is represented using an adjacency list - an easy way to do it in Python is to use a dictionary data structure. Each vertex has a list of its adjacent nodes stored.
• Line 11: visited is a set that is used to keep track of visited nodes.
• Line 21: The dfs function is called and is passed the visited set, the graph in the form of a dictionary, and A, which is the starting node.
• Lines 13-18: dfs follows the algorithm described above:
1. It first checks if the current node is unvisited - if yes, it is appended in the visited set.
2. Then for each neighbor of the current node, the dfs function is invoked again.
3. The base case is invoked when all the nodes are visited. The function then returns.

## Time Complexity

Since all the nodes and vertices are visited, the average time complexity for DFS on a graph is $O(V + E)$, where $V$ is the number of vertices and $E$ is the number of edges. In case of DFS on a tree, the time complexity is $O(V)$, where $V$ is the number of nodes.

Note: We say average time complexity because a set’s in operation has an average time complexity of $O(1)$. If we used a list, the complexity would be higher.

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