133. Clone Graph

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.

Example:

Input:
{"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1}

Explanation:
Node 1's value is 1, and it has two neighbors: Node 2 and 4.
Node 2's value is 2, and it has two neighbors: Node 1 and 3.
Node 3's value is 3, and it has two neighbors: Node 2 and 4.
Node 4's value is 4, and it has two neighbors: Node 1 and 3.

Note:

  1. The number of nodes will be between 1 and 100.

  2. The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.

  3. Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.

  4. You must return the copy of the given node as a reference to the cloned graph.

解题要点:

学习大神解法,使用DFS。

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, neighbors):
        self.val = val
        self.neighbors = neighbors
"""
class Solution(object):
    def cloneGraph(self, node):
        """
        :type node: Node
        :rtype: Node
        """
        mydict = {}
        def dfs(node):
            if node and node not in mydict:
                clone = Node(node.val, node.neighbors)
                mydict[node] = clone
                clone.neighbors = [mydict.get(n) or dfs(n) for n in node.neighbors]
                return clone
        return dfs(node)

Last updated

Was this helpful?