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],
]

confused what

"{1,#,2,3}"

means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as

"{1,2,3,#,#,4,#,#,5}"

.
思路：此题和Binary Tree Level Order Traversal是类似的，就是输出的顺序改变了，在上题的基础之上，我将顺序颠倒输出就可以了。

/**
* 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) {
vector<vector<int> > result;
if(root==NULL)
return result;
queue<TreeNode *> pTree;
vector<int> data;
pTree.push(root);
int count=1;
while(!pTree.empty())
{
data.clear();
int temp_count=0;
for(int i=0;i<count;i++)
{
TreeNode *tn=pTree.front();
pTree.pop();
data.push_back(tn->val);
if(tn->left)
{
pTree.push(tn->left);
temp_count++;
}
if(tn->right)
{
pTree.push(tn->right);
temp_count++;
}
}
count=temp_count;
result.push_back(data);
}
vector<vector<int> > ret;
vector<vector<int> >::iterator iter=result.end()-1;
for(;iter>=result.begin();iter--)
{
ret.push_back(*iter);
}
return ret;
}
};