107. Binary Tree Level Order Traversal II（二叉树的层序遍历II）

题目描述

方法思路

class Solution {
    //Runtime: 1 ms, faster than 92.29%
    //Memory Usage: 37.2 MB, less than 74.03%
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new LinkedList<>();
        if(root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            Queue<TreeNode> temp = new LinkedList<>();
            List<Integer> list = new ArrayList<>();
            while(!queue.isEmpty()){
                TreeNode node = queue.poll();
                list.add(node.val);
                if(node.left != null)
                    temp.offer(node.left);
                if(node.right != null)
                    temp.offer(node.right);
            }
            queue = temp;
            res.add(0, list);//不懂为什么不能直接res.addFirst(list);
        }
       /* List<List<Integer>> res2 = new LinkedList<>();
        for(int i = res.size() - 1; i >= 0; i--){
            res2.add(res.get(i));
        }
        return res2;*/
        return res;
    }
}