题目：
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]
]
代码：
/**
* Definition for a binary tree node.
* 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> > ret;
if (!root) return ret;
vector<int> tmp_ret;
deque<TreeNode *> currLevel, nextLevel;
currLevel.push_back(root);
while ( !currLevel.empty() )
{
while ( !currLevel.empty() )
{
TreeNode * tmp = currLevel.front();
currLevel.pop_front();
tmp_ret.push_back(tmp->val);
if ( tmp->left ) nextLevel.push_back(tmp->left);
if ( tmp->right ) nextLevel.push_back(tmp->right);
}
ret.push_back(tmp_ret);
tmp_ret.clear();
std::swap(currLevel, nextLevel);
}
std::reverse(ret.begin(), ret.end());
return ret;
}
};
tips：
在Binary Tree Level Order Traversal的基础上加一个reverse即可。
============================================
第二次过这道题，直接用reverse的路子了。
/**
* Definition for a binary tree node.
* 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> > ret;
queue<TreeNode*> curr;
queue<TreeNode*> next;
if ( root ) curr.push(root);
while ( !curr.empty() )
{
vector<int> tmp;
while ( !curr.empty() )
{
tmp.push_back(curr.front()->val);
if ( curr.front()->left ) next.push(curr.front()->left);
if ( curr.front()->right ) next.push(curr.front()->right);
curr.pop();
}
ret.push_back(tmp);
std::swap(next, curr);
}
std::reverse(ret.begin(), ret.end());
return ret;
}
};