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这题第一眼看到想到的是BFS。接着想到我需要一个queue来做BFS。之后，为了能一一对应的复制，还用到一个Map。对应新旧node。
BUG： 就是我在复制新child node之前，就把旧child node添加到新的node的neighbors中了。
1 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { 2 if (node == null) { 3 return null; 4 } 5 Queue<UndirectedGraphNode> q = new LinkedList<UndirectedGraphNode>(); 6 Map<UndirectedGraphNode,UndirectedGraphNode> map = new HashMap<UndirectedGraphNode,UndirectedGraphNode>(); 7 UndirectedGraphNode nodeCopy = new UndirectedGraphNode(node.label); 8 q.add(node); 9 map.put(node, nodeCopy); 10 while (!q.isEmpty()) { 11 UndirectedGraphNode cur = q.poll(); 12 UndirectedGraphNode curCopy = map.get(cur); 13 for (UndirectedGraphNode child : cur.neighbors) { 14 if (!map.containsKey(child)) { 15 UndirectedGraphNode childCopy = new UndirectedGraphNode(child.label); 16 map.put(child, childCopy); 17 q.offer(child); 18 } 19 //BUG：先复制child node 20 curCopy.neighbors.add(map.get(child)); 21 } 22 } 23 return map.get(node); 24 }
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原文地址：http://www.cnblogs.com/gonuts/p/4391970.html
