102 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,null,null,15,7]

,

3
/ \
9  20
/  \
15   7

return its level order traversal as:

[
[3],
[9,20],
[15,7]
]

BFS模板题。
粗暴了，直接两个队列，一个记录node，一个记录深度

public List<List<Integer>> levelOrder(TreeNode root)
{
List<List<Integer>> retlist=new ArrayList<>();
if(root==null)
return retlist;

ArrayDeque<TreeNode> deque=new ArrayDeque<>();
ArrayDeque<Integer> leveldeq=new ArrayDeque<>();
deque.add(root);
leveldeq.add(1);
int prelevel=1;
ArrayList<Integer> arraylist=new ArrayList<>();
while(!deque.isEmpty())
{
TreeNode t=deque.poll();
int level=leveldeq.poll();
if(level>prelevel)
{
retlist.add(new ArrayList<>(arraylist));
arraylist.clear();
}
arraylist.add(t.val);
if(t.left!=null)
{
deque.add(t.left);
leveldeq.add(level+1);
}
if(t.right!=null)
{
deque.add(t.right);
leveldeq.add(level+1);
}
prelevel=level;
}
retlist.add(arraylist);
return retlist;
}

update 2016.7.20

深度不用额外使用队列来记录，设置计数值，每次出队就减少计数值，每当减到0表明本层已经遍历完毕，下一层的元素数目就是当前队列的size，把计数值重置为当前队列size即可

public List<List<Integer>> levelOrder(TreeNode root)
{
List<List<Integer>> retlist=new ArrayList<>();
if(root==null)
return retlist;

ArrayDeque<TreeNode> deque=new ArrayDeque<>();
deque.add(root);
int level=1;
ArrayList<Integer> arraylist=new ArrayList<>();
while(!deque.isEmpty())
{
TreeNode t=deque.poll();

if(t.left!=null)
deque.add(t.left);

if(t.right!=null)
deque.add(t.right);

level--;
arraylist.add(t.val);

if(level==0)
{
retlist.add(new ArrayList<>(arraylist));
arraylist.clear();
level=deque.size();
}

}
return retlist;
}