version at: 2022-04-20

/**
 * 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;
 *     }
 * }
 */
/**
1. 类似广搜, 用队列维护下, 记录当前层的数量, pop时把下一层都放到队列里
*/

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        List<List<Integer>> results = new ArrayList<>();

        if (root != null) queue.add(root);

        while (!queue.isEmpty()) {
            List<Integer> result = new ArrayList<>();
            int count = queue.size();

            for (int i = 0; i < count ;i++) {
                TreeNode node = queue.poll();

                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);

                result.add(node.val);
            }

            results.add(result);
        }

        return results;
    }
}

version at: 2020-02-04

/*
1. 按层遍历树，主要是在宽搜的时候分辨出各层的值
2. 记录个数 或 用滚动队列都可以， 记录个数更剩空间
3，记得判空

*/


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

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

        Queue<TreeNode>[] qs = new Queue[2];
        qs[0] = new LinkedList<>();
        qs[1] = new LinkedList<>();

        int cur = 0;
        qs[cur].add(root);

        // while (!qs[0].isEmpty() && !qs[1].isEmpty()){
        while (!qs[0].isEmpty() || !qs[1].isEmpty()){
            List<Integer> list = new ArrayList<>();
            while(!qs[cur].isEmpty()){
                TreeNode node = qs[cur].poll();
                list.add(node.val);
                if (node.left != null) qs[cur^1].add(node.left);
                if (node.right != null) qs[cur^1].add(node.right);
            }

            cur ^=1;
            res.add(list);
        }
        return res;
    }
}