LeetCode-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]
]

重点是如何将每层隔开，使用广度优先遍历，时间复杂度为O(n):

 public List<List<Integer>> levelOrder(TreeNode root) {//树 BFS my
        Queue<TreeNode> queue = new LinkedList<>();
        List<List<Integer>> re = new ArrayList<>();
        if(null!=root){
            queue.add(root);
            queue.add(null);
        }
        while(!queue.isEmpty()&&null!=queue.peek()){//本方法使用null隔开每一层，也可以记录每一层的个数
            List<Integer> list = new ArrayList<>();
            TreeNode cur = queue.poll();
            while(null!=cur){
                list.add(cur.val);
                if(null!=cur.left){
                    queue.add(cur.left);
                }
                if(null!=cur.right){
                    queue.add(cur.right);
                }
                cur=queue.poll();
            }
            queue.add(null);
            if(!list.isEmpty()){
                re.add(list);
            }
        }
        return re;
    }

也可以使用DFS完成

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