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]
]

我估计下面这个方法不能算是最优的，大牛肯定有更简单的方法。

/**
* 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> > results;
queue<TreeNode *> onelevel;
if(root == NULL)
return results;
onelevel.push(root);
while(!onelevel.empty()) {
vector<int> onelevelvals;
queue<TreeNode *> newlevel;
while(!onelevel.empty()) {
TreeNode *front = onelevel.front();
onelevelvals.push_back(front->val);
if(front->left != NULL)
newlevel.push(front->left);
if(front->right != NULL)
newlevel.push(front->right);
onelevel.pop();
}
results.push_back(onelevelvals);
onelevel = newlevel;
}
reverse(results.begin(), results.end());
return results;
}
};