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,null,null,15,7],

    3
/ \
9 20
/ \
15 7

return its level order traversal as:

[
[3],
[9,20],
[15,7]
] 层序遍历 利用queue
 class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
if(root==null) return res;
queue.add(root);
while (!queue.isEmpty()) {
int levlnum = queue.size();
List<Integer> res_temp = new ArrayList<Integer>();
for (int i = 0;i<levlnum ;i++ ){ //把每层的左右节点都保存到queue里
//并讲当层的值从queue里弹出,加到res_temp 中
if(queue.peek().left!=null) queue.add(queue.peek().left);
if(queue.peek().right!=null) queue.add(queue.peek().right); res_temp.add(queue.poll().val);
}
res.add(res_temp);
}
return res;
}
}

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