leetcode--Binary Tree Level Order Traversal II

[code]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],

]

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]/**

* 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

vector<vector<int>> result;

if(!root)return result;

queue<TreeNode *>q;

int num_curLayer=1;

int num_nextLayer=0;

q.push(root);

vector<int>curVec;

while(!q.empty())

{

    curVec.clear();   

    while(num_curLayer>0)

    {

TreeNode *cur = q.front();

q.pop();num_curLayer--;

curVec.push_back(cur->val);

if(cur->left){q.push(cur->left);num_nextLayer++;}

if(cur->right){q.push(cur->right);num_nextLayer++;}

}

    result.push_back(curVec);

    num_curLayer=num_nextLayer;

    num_nextLayer=0;

}

std::reverse(result.begin(),result.end());

return result;

}

};