Leetcode: Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

Best approach:

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<> ();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> row = new ArrayList<>();
            for (int i = 0; i < size; i ++) {
                TreeNode node = queue.poll();
                row.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            res.add(row);
        }
        return res;
    }
}

把树看成一个有向图，进行BFS。BST的通常做法都是维护一个队列。这道题的难度在于，如何在队列不断的入队出队操作中，确定哪些node是一个层次的。想了很久没什么好办法，参考了网上的做法，发现他在维护两个数：一个是父节点所在层次的节点在当前队列中的数目ParentNumInQ，另一个是子节点所在层次的节点在当前队列中的数目ChildNumInQ。初始数值为1和0. 每次父节点出队，ParentNumInQ--，每次子节点入队，ParentNumInQ++。一旦ParentNumInQ减至0，说明当前父节点层次已全部出队，全部被存入LinkedList中，这就得到了一层的所有节点。接下来需要一些更新，ParentNumInQ = ChildNumInQ, ChildNumInQ = 0开始考虑下一层。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>> ();
        if (root == null) return lists;
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int ParentNumInQ = 1;
        int ChildNumInQ = 0;
        ArrayList<Integer> list = new ArrayList<Integer>();
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            list.add(cur.val);
            ParentNumInQ--;
            if (cur.left != null) {
                queue.add(cur.left);
                ChildNumInQ++;
            }
            if (cur.right != null) {
                queue.add(cur.right);
                ChildNumInQ++;
            }
            if (ParentNumInQ == 0) {
                ParentNumInQ = ChildNumInQ;
                ChildNumInQ = 0;
                lists.add(list);
                list = new ArrayList<Integer>();
            }
        }
        return lists;
    }
}