Binary Tree Level Order Traversal II

Problem:

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

Solution:
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:

// Use reverse(sol.begin(), sol.end()) instead of following.
    /*
    void reverse(vector<vector<int> > &sol) {
        int size = sol.size();
        int start = 0;
        int end = size - 1;
        while (start < end) {
            vector<int> temp;
            temp = sol[start];
            sol[start] = sol[end];
            sol[end] = temp;
            start++;
            end--;
        }
    }
    */
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > sol;
        
        if (root == NULL)
            return sol;
        vector<TreeNode *> prev, curr;
        
        prev.push_back(root);
        while (prev.size() != 0) {
            int size = prev.size();
            vector<int> curr_int;
            for (int i = 0; i < size; i++) {
                curr_int.push_back(prev[i] -> val);
                if (prev[i] -> left)
                    curr.push_back(prev[i] -> left);
                if (prev[i] -> right)
                    curr.push_back(prev[i] -> right);
            }
            prev = curr;
            curr.clear();
            sol.push_back(curr_int);
        }
        reverse(sol.begin(), sol.end());
        return sol;
    }
};