Leetcode 107. Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]

题意：按层级倒序遍历二叉树。

思路：这道题是102题的followup，要求从正序遍历变为倒序，所以只需要在102题的解法上加上一行reverse代码即可。

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

    Queue q = new LinkedList<>();
    q.offer(root);
    while (!q.isEmpty()) {
        int size = q.size();
        List list = new ArrayList<>();
        for (int i = 0; i < size; i++) {
            TreeNode cur = q.poll();
            list.add(cur.val);
            if (cur.left != null) {
                q.offer(cur.left);
            }
            if (cur.right != null) {
                q.offer(cur.right);
            }
        }
        res.add(list);
    }
    //翻转，逆序输出
    Collections.reverse(res);

    return res;
}