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：

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10  struct node
11  {
12      TreeNode *val;
13      int dep;
14  };
15 class Solution {
16 public:
17     vector<vector<int>>  result;
18     vector<int>  temp;
19     vector<vector<int> > levelOrderBottom(TreeNode *root) {
20         doit(root,0);
21         reverse(result.begin(),result.end());
22         return result;
23     }
24
25     void doit(TreeNode *root,int level)
26     {
27         if(root==NULL)  return;
28         else
29         {
30             if(level==result.size())
31             {
32                  vector<int> v;
33                  result.push_back(v);
34             }
35             result[level].push_back(root->val);
36             doit(root->left,level+1);
37             doit(root->right,level+1);
38         }
39     }
40 };