[leetcode]Binary Tree Level Order Traversal II
2013-07-02 11:49:05

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],
] /**

```* 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
stack<vector<int>> s;
if(!root) return vector<vector<int>>();

queue<TreeNode*> q1,q2;
q1.push(root);

TreeNode *cur;
vector<int> tmp;

while(!q1.empty()){
tmp.clear();
while(!q1.empty()){
cur = q1.front();
q1.pop();

tmp.push_back(cur -> val);
if(cur -> left) q2.push(cur -> left);
if(cur -> right) q2.push(cur -> right);
}
s.push(tmp);
swap(q1, q2);
}

vector<int> curVec;
vector<vector<int>> result;
while(!s.empty()){
curVec = s.top();
s.pop();
result.push_back(curVec);
}
return result;

}

};

/**
* 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
stack<vector<int>> s;
if(!root) return vector<vector<int>>();

queue<TreeNode*> q1,q2;
q1.push(root);

TreeNode *cur;
vector<int> tmp;

while(!q1.empty()){
tmp.clear();
while(!q1.empty()){
cur = q1.front();
q1.pop();

tmp.push_back(cur -> val);
if(cur -> left) q2.push(cur -> left);
if(cur -> right) q2.push(cur -> right);
}
s.push(tmp);
swap(q1, q2);
}

vector<int> curVec;
vector<vector<int>> result;
while(!s.empty()){
curVec = s.top();
s.pop();
result.push_back(curVec);
}
return result;

}

};

```