Problem

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]
]

思路

• queue, when it is not empty, add left and right to the queue

    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();

        if (root == null) return result;

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            ArrayList<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode curt = queue.poll();
                level.add(curt.val);
                if (curt.left != null) {
                    queue.offer(curt.left);
                }
                if (curt.right != null) {
                    queue.offer(curt.right);
                }
            }
            result.add(level);
        }
        return result;
    }