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],
]

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}"

.

/**
* 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
queue<TreeNode*> q;
vector<vector<int> >ret;
if(root==NULL)return ret;
q.push(root);
int level=0;
ret.resize(level+1);
int count=1;
int nextCount=0;
while(q.size()>0){
if(count==0){
level++;
ret.resize(level+1);
count=nextCount;
nextCount=0;
}
TreeNode* current=q.front();
ret[level].push_back(current->val);
q.pop();
if(current->left!=NULL){
q.push(current->left);
nextCount++;
}
if(current->right!=NULL){
q.push(current->right);
nextCount++;
}
count--;

}
int half=ret.size()/2;
for(int i=0;i<half;i++){
vector<int> temp=ret[i];
ret[i]=ret[ret.size()-i-1];
ret[ret.size()-i-1]=temp;
}
return ret;
}
};