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 val (int) and a list (List[Node]) of its neighbors.

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

Test case format:

For simplicity sake, each node’s value is the same as the node’s index (1-indexed). For example, the first node with val = 1, the second node with val = 2, and so on. The graph is represented in the test case using an adjacency list.

Adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

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).

Example 2:

Input: adjList = [[]]
Output: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:

Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.

Example 4:

Input: adjList = [[2],[1]]
Output: [[2],[1]]

Constraints:

  • 1 <= Node.val <= 100
  • Node.val is unique for each node.
  • Number of Nodes will not exceed 100.
  • There is no repeated edges and no self-loops in the graph.
  • The Graph is connected and all nodes can be visited starting from the given node.

Recursive DFS Solution [w/ Diagram]

Review:

This one is very similar to Leetcode #138 Copy List with Random Pointer

The key idea behind is the same:

  1. Create deep copy for each node, and maintain the mapping relation
    image
  2. Rebuild node linkage by mapping relation
    image

Hint:

Stage_1:

Create deep copy for each node in DFS
Maintain a mapping relation (i.e., hash map or dictionary) between original ones and deep-copy ones.


Stage_2:

Rebuild neighbor list in DFS with mapping relation.


Implementation:

class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':
        
        # key: memory id of original node
        # value: corresponding deep copy node
        mapping = {}
        
        # -----------------------------------------
        def helper( node: 'Node' ) -> 'Node':
            
            if not node:
    
                # empty node's deep copy is still empty node
                return node
            
            elif id(node) in mapping:
                
                # current node already has deep copy
                return mapping[ id(node) ]
            
            # create deep copy for current node
            mapping[ id(node) ] = Node( val=node.val, neighbors=[] )
            
            for original_neighbor in node.neighbors:
                # update neighbor list for current node
                mapping[ id(node) ].neighbors.append( helper(original_neighbor) )
            
            return  mapping[ id(node) ]
        
        # -----------------------------------------
        return helper( node )

RECURSIVE DFS

"""
# 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 []
"""

class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':
        def dfs(curr):
            if not curr: return curr
            
            if curr in visited:
                return visited[curr]
                
            
            clone = Node(curr.val,[])
            
            visited[curr] = clone
            
            if curr.neighbors:
                clone.neighbors = [dfs(n) for n in curr.neighbors]
            
            return clone
        
        visited = {}
        return dfs(node)

Python BFS

"""
# 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 []
"""

class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':
        if not node:	
            return node
        
        visited = {}
        queue = deque([node])
        visited[node] = Node(node.val, [])
        
        while queue:
            n = queue.popleft()
            
            for neighbor in n.neighbors:
                if neighbor not in visited:
                    visited[neighbor] = Node(neighbor.val, [])
                    queue.append(neighbor)
                visited[n].neighbors.append(visited[neighbor])
                
        return visited[node]

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