LeetCode 107. Binary Tree Level Order Traversal II

description:
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]
]
题目并不复杂，只是在binary tree level order traversal 的基础之上进行的一个follow up

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> levelOrderBottom(TreeNode root) {        List<List<Integer>> result = new ArrayList<>();        if (root == null) {            return result;        }        Queue<TreeNode> queue = new LinkedList<>();        queue.add(root);        while(!queue.isEmpty()) {            List<Integer> level = new ArrayList<>();            int len = queue.size();            for (int i = 0; i < len; i++) {                TreeNode node = queue.poll();                level.add(node.val);                if (node.left != null) {                    queue.offer(node.left);                }                if (node.right != null) {                    queue.offer(node.right);                }            }            result.add(level);        }        List<List<Integer>> results = new ArrayList<>();        for(int i = result.size() - 1; i >= 0; i--) {            results.add(result.get(i));        }        return results;    }}