1. 题目描述（简单难度）

[success] 102. 二叉树的层序遍历

2. 解法一：使用队列进行层序遍历

先巩固一下层序遍历输出

class Solution {
    public static void main(String[] args) {
        TreeNode t1 = new TreeNode(3);
        TreeNode t2 = new TreeNode(9);
        TreeNode t3 = new TreeNode(20);
        TreeNode t4 = new TreeNode(15);
        TreeNode t5 = new TreeNode(7);

        t1.left = t2;
        t1.right = t3;
        t2.left = t4;
        t2.right = t5;

        List<Integer> resp = levelOrder(t1);
        resp.forEach(System.out::println);
    }

    public static List<Integer> levelOrder(TreeNode root) {
        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        List<Integer> res = new ArrayList<>();
        while (!deque.isEmpty()) {
            TreeNode levelRoot = deque.poll();
            res.add(levelRoot.val);
            if (levelRoot.left != null) {
                deque.offer(levelRoot.left);
            }
            if (levelRoot.right != null) {
                deque.offer(levelRoot.right);
            }
        }
        return res;
    }
}

本题解题思路： BFS

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (null == root) {
            return new ArrayList<>();
        }
        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        List<List<Integer>> resp = new ArrayList<>();
        while (!deque.isEmpty()) {
            List<Integer> ans = new ArrayList<>();
            int level = deque.size();
            for (int i = 0; i < level; i++) {
                TreeNode poll = deque.poll();
                ans.add(poll.val);
                if (poll.left != null) {
                    deque.offer(poll.left);
                }
                if (poll.right != null) {
                    deque.offer(poll.right);
                }
            }
            resp.add(ans);
        }
        return resp;
    }
}