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],
]

还是二叉树的层次遍历，之后返回结果要从底层开始，偷了个懒，直接把从顶层开始的结果倒过来，哈哈哈哈~~~~

/**
* 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) {
queue<TreeNode*>q1;
vector<vector<int>>ovec,ret;
vector<int>ivec;
TreeNode *temp = root;
if(!temp)
return ret;
q1.push(temp);
do
{
queue<TreeNode*>q2;;
while (!q1.empty())
{
temp = q1.front();
ivec.push_back(temp->val);
q1.pop();
if (temp->left)
q2.push(temp->left);
if (temp->right)
q2.push(temp->right);
}
ovec.push_back(ivec);
ivec.clear();
q1 = q2;
} while (!q1.empty());
for (vector<vector<int>>::size_type i = 0; i < ovec.size(); ++i)
ret.push_back(ovec[ovec.size() - 1 - i]);
return ret;
}
};