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).

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]
]

分析：

用两个ArrayList,一个保存当前这行所有的Node, 然后另一个数组保存下一行所有的nodes.

 /**
  * Definition of TreeNode:
  * public class TreeNode {
  *     public int val;
  *     public TreeNode left, right;
  *     public TreeNode(int val) {
  *         this.val = val;
  *         this.left = this.right = null;
  *     }
  * }
  */

 public class Solution {
     /**
      * @param root: The root of binary tree.
      * @return: Level order a list of lists of integer
      */
     public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
         ArrayList<ArrayList<Integer>> allLists = new ArrayList<>();
         if (root == null) return allLists;

         ArrayList<Integer> nums = new ArrayList<>();
         ArrayList<TreeNode> list1 = new ArrayList<>();
         ArrayList<TreeNode> list2 = new ArrayList<>();

         list1.add(root);

         while(list1.size() != ) {
             for (TreeNode node : list1) {
                 nums.add(node.val);
                 if (node.left != null) list2.add(node.left);
                 if (node.right != null) list2.add(node.right);
             }
             list1.clear();
             allLists.add(new ArrayList<Integer>(nums));
             nums.clear();
             ArrayList<TreeNode> temp = list1;
             list1 = list2;
             list2 = temp;
         }
         return allLists;
     }
 }