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LeetCode

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Problem

Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label (int) and a list (List[UndirectedGraphNode]) of its neighbors. There is an edge between the given node and each of the nodes in its neighbors.

OJ’s undirected graph serialization (so you can understand error output): Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2. Second node is labeled as 1. Connect node 1 to node 2. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:


       1
      / \
     /   \
    0 --- 2
         / \
         \_/

Note: The information about the tree serialization is only meant so that you can understand error output if you get a wrong answer. You don’t need to understand the serialization to solve the problem.

Code

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> neighbors;

    public Node() {}

    public Node(int _val,List<Node> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
};
*/
class Solution {
    public Node cloneGraph(Node node) {
        if(node == null) return node;
        HashMap<Node, Node> map = new HashMap<>();
        // 两步: 先copy点, 再copy边
        copyNode(node, map);
        copyNei(node, map, new HashSet<Node>());

        return map.get(node);
    }

    private void copyNode(Node node, HashMap<Node,Node> map){
        if(map.containsKey(node)) return;
        map.put(node, new Node(node.val, new ArrayList<>()));

        for(Node nei : node.neighbors){
            copyNode(nei, map);
        }
    }

    private void copyNei(Node node, HashMap<Node, Node> map, HashSet<Node> visited){
        if(visited.contains(node)) return;
        visited.add(node);

        for(Node nei : node.neighbors){
            map.get(node).neighbors.add(map.get(nei));
            copyNei(nei, map, visited);
        }
    }
}