Binary Tree Level Order Traversal II <leetcode>

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]
]

思路：本来想到用分支限界，但是代码写的很复杂，没有控制住，后来在网上看到一个方法，很简单，直接遍历，有些代码挺巧妙的，代码如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 struct node
 {
     TreeNode *val;
     int dep;
 };
class Solution {
public:
    vector<vector<int>>  result;
    vector<int>  temp;
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        doit(root,0);
        reverse(result.begin(),result.end());
        return result;
    }
    
    void doit(TreeNode *root,int level)
    {
        if(root==NULL)  return;
        else
        {
            if(level==result.size())
            {
                 vector<int> v;
                 result.push_back(v);
            }
            result[level].push_back(root->val);
            doit(root->left,level+1);
            doit(root->right,level+1);
        }
    }
};