Binary Tree Level Order Traversal

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]
]

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> container = new ArrayList<List<Integer>>();
        if(null == root)
            return container;
        //List<TreeNode> theSameLevel = new ArrayList<TreeNode>();//存放同一层节点
        Queue<TreeNode> theSameLevel = new LinkedList<TreeNode>();
        theSameLevel.add(root);
        while(!theSameLevel.isEmpty()){//theSameLevel不为空
            List<Integer> oneLevel = new ArrayList<Integer>();
            Queue<TreeNode> temp = new LinkedList<TreeNode>();//暂存同一层的结点
            while(!theSameLevel.isEmpty()){
                TreeNode cur = theSameLevel.remove();
                oneLevel.add(cur.val);
                if(null != cur.left) 
                    temp.add(cur.left);
                if(null != cur.right)
                    temp.add(cur.right);
            }
            theSameLevel = temp;
            container.add(oneLevel);
        }
        
        return container;
    }
}