[leetcode]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,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its bottom-up level order traversal as:
[
  [15,7]
  [9,20],
  [3],
]

偷了一下懒，将上一题的结果反过来就是 。。。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > answers;
        if(root == NULL)
            return answers;
        queue<TreeNode *> * nodes = new queue<TreeNode *>();
        vector<int> * v = NULL;
        TreeNode * leftFirst;
        nodes->push(root);
        leftFirst = root;
        bool hasLeftFirst = true;
        while(!nodes->empty())
        {
            TreeNode * tmp = nodes->front();
            nodes->pop();
            if(tmp == leftFirst)
            {
                if(v != NULL)
                    answers.push_back(*v);
                v = new vector<int>();
                hasLeftFirst = false;
            }
            v->push_back(tmp->val);
            if(tmp->left != NULL)
            {
                nodes->push(tmp->left);
                if(hasLeftFirst == false)
                {
                    hasLeftFirst = true;
                    leftFirst = tmp->left;
                }
            }
            if(tmp->right != NULL)
            {
                nodes->push(tmp->right);
                if(hasLeftFirst == false)
                {
                    hasLeftFirst = true;
                    leftFirst = tmp->right;
                }
            }
        }
        if(v != NULL)
            answers.push_back(*v);
        vector<vector<int> > answers1;
        for(int i = answers.size()-1;i>=0;i--)
            answers1.push_back(answers.at(i));
        return answers1;     
    }
};