将伪代码调整为java实现以查找特里结构中最长的单词

Question

参考这个问题，我问：How to find the longest word in a trie?

我在实现答案中给出的伪代码时遇到了麻烦。

findLongest(trie):
 //first do a BFS and find the "last node"
 queue <- []
 queue.add(trie.root)
 last <- nil
 map <- empty map
while (not queue.empty()):
 curr <- queue.pop()
 for each son of curr:
    queue.add(son)
    map.put(son,curr) //marking curr as the parent of son
 last <- curr
//in here, last indicate the leaf of the longest word
//Now, go up the trie and find the actual path/string
curr <- last
str = ""
while (curr != nil):
      str = curr + str //we go from end to start   
    curr = map.get(curr)
return str

这就是我的方法

public static String longestWord (DTN d) {
  Queue<DTN> holding = new ArrayQueue<DTN>();
  holding.add(d);
  DTN last = null;
  Map<DTN,DTN> test = new ArrayMap<DTN,DTN>();
  DTN curr;
  while (!holding.isEmpty()) {
       curr = holding.remove();

      for (Map.Entry<String, DTN> e : curr.children.entries()) {
          holding.add(curr.children.get(e));
          test.put(curr.children.get(e), curr);
      }
          last = curr;

      }
  curr = last;
  String str = "";
  while (curr != null) {
      str = curr + str;
      curr = test.get(curr);

  }
  return str;

  }

我在：

收到NullPointerException

 for (Map.Entry<String, DTN> e : curr.children.entries())

如何找到并修复方法的NullPointerException的原因，以便它返回trie中最长的单词？

Answer 1

确保curr.children.entries()不是空值。也许，如果该节点没有子节点，DTN将返回空值。这会导致您的NullPointerException。

在开始迭代之前尝试快速检查。

if(curr.children.entries() != null)
{
    //It's safe, so procede with going deeper into the trie.
}

Answer 2

除了@ Clark的答案之外，请确保在解除引用之前curr.children不为空。