原题出处：https://leetcode.cn/leetbook/read/top-interview-questions-easy/xnldjj/

解法一：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null) {
            return new ArrayList();
        }
        Queue<TreeNode> queue = new LinkedList();
        List<List<Integer>> res = new ArrayList();
        queue.add(root);
        while(!queue.isEmpty()) {
            int len = queue.size();
            List<Integer> list = new ArrayList();
            for (int i = 0 ; i < len; i++) {
                TreeNode treeQueue = queue.poll();
                list.add(treeQueue.val);
                if (treeQueue.left != null) {
                    queue.add(treeQueue.left);
                }
                if (treeQueue.right != null) {
                    queue.add(treeQueue.right);
                }
            }
            res.add(list);
        }
        return res;
    }
}

思路：BFS，广度优先搜索，从根节点开始遍历之后将它的左子树和右子树进行进行依次遍历即可。

解法二：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList();
        levelFun(res,root,0);
        return res;
    }

    public void levelFun(List<List<Integer>> res,TreeNode root,int level) {
        if (root == null) {
            return;
        }
        if (level >= res.size()) {
            res.add(new ArrayList());
        }
        res.get(level).add(root.val);
        levelFun(res,root.left,level+1);
        levelFun(res,root.right,level+1);
    }

}

思路：DFS，深度优先，遍历左子树与右子树的深度优先