Tree - Clone Graph

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Difficulty : Easy

DFS, Map

Problem

Given a reference of a node in a connected undirected graph. Return a deep copy (clone) of the graph.

Each node in the graph contains a value (int) and a list (List[Node]) of its neighbors.

class Node {
    public int val;
    public List<Node> neighbors;
} 

Example 1:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

Solution

High Level Explanation

1. Very simple and basic DFS problem.
2. Only new line is the creation of new Node().
3. The Map is a mapping between old (key) and new (val) node.

Maintain a visited map in the main function. We are using a dict object and not a set as we need more than just to track the nodes. We also want to get the newly created duplicate node for the original node whenever returning a visit to the node.

visited = {}

Now we can start our dfs() function. If we are revisiting a already visited node, just return the newly created node.

def dfs(node):
  if node in visited:
    return visited[node]

Otherwise, create a new node by copying existing node.

dup_node = Node(node.val)
visited[node] = dup_node

Traverse all the neighbors by running dfs() on each neighbor. This way the dfs() function will always return the newly created duplicate node.

for neighbor in node.neighbors:
  dup_neighbor = dfs(neighbor)

Now add the dup_neighbor as new neighbor to current node.

  dup_node.neighbors.append(dup_neighbor)

Finally, return the new node.

return dup_node

At the end, call the dfs() function.

return dfs(node) if node else None

Final Code

Here is the full code.

"""
# Definition for a Node.
class Node:
    def __init__(self, val = 0, neighbors = None):
        self.val = val
        self.neighbors = neighbors if neighbors is not None else []
"""
        
def clone_graph(node):
    visited = {}

    def dfs(node):
        if node in visited:
            return visited[node]

        dup_node = Node(node.val)
        visited[node]= dup_node

        for neighbor in node.neighbors:
            dup_node.neighbors.append(dfs(neighbor))

        return dup_node
    return dfs(node) if node else None