364. Nested List Weight Sum II[没有解出]

其实本题并不难，没有解出主要是题干没有写清楚，不知道是level还是height，
后来搞明白[[b], a, [c, d]]，在这个例子中，b是在level 2的，而不是level 1，即便他是leaf node。
这样就是一个很简单的用BFS的问题。
代码如下:

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *     // Constructor initializes an empty nested list.
 *     public NestedInteger();
 *
 *     // Constructor initializes a single integer.
 *     public NestedInteger(int value);
 *
 *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
 *     public boolean isInteger();
 *
 *     // @return the single integer that this NestedInteger holds, if it holds a single integer
 *     // Return null if this NestedInteger holds a nested list
 *     public Integer getInteger();
 *
 *     // Set this NestedInteger to hold a single integer.
 *     public void setInteger(int value);
 *
 *     // Set this NestedInteger to hold a nested list and adds a nested integer to it.
 *     public void add(NestedInteger ni);
 *
 *     // @return the nested list that this NestedInteger holds, if it holds a nested list
 *     // Return null if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
class Solution {
  public int depthSumInverse(List<NestedInteger> nestedList) {
    if (nestedList == null || nestedList.isEmpty()) {
      return 0;
    }
    Queue<NestedInteger> queue = new LinkedList<>();
    queue.addAll(nestedList);
    int preSum = 0;
    int sum = 0;
    while (!queue.isEmpty()) {
      int levelSum = 0;
      int sz = queue.size();
      for (int i = 0; i < sz; ++i) {
        NestedInteger ni = queue.poll();
        if (ni.isInteger()) {
          levelSum += ni.getInteger();
        }
        else {
          queue.addAll(ni.getList());
        }
      }
      preSum += levelSum;
      sum += preSum;
    }
    return sum;
  }
}