很簡單的一道題，將前一道題的結果數組進行一次反轉即可。

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:void order(TreeNode* cur, vector<vector<int>>& result, int depth){if(cur == nullptr){return;}if(result.size() == depth){result.push_back(vector<int>());}result[depth].push_back(cur->val);order(cur->left, result, depth + 1);order(cur->right, result, depth + 1);}vector<vector<int>> levelOrderBottom(TreeNode* root) {vector<vector<int>> result;int depth = 0;order(root, result, depth);reverse(result.begin(), result.end());return result;}
};