【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}"

.
code: 1A 36ms

/**
* 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) {
// Note: The Solution object is instantiated only once and is reused by each test case.
vector<vector<int> > res;
if(root == NULL)
return res;
queue<TreeNode *> qu;
qu.push(root);
qu.push(NULL);
vector<int> onelevel;
while(true)
{
TreeNode *cur = qu.front();
qu.pop();
if(cur == NULL)
{
res.push_back(onelevel);
onelevel.clear();
if(qu.empty())
break;
qu.push(NULL);
}
else
{
onelevel.push_back(cur->val);
if(cur->left)
qu.push(cur->left);
if(cur->right)
qu.push(cur->right);
}

}
reverse(res.begin(),res.end());
return res;
}
};