leetCode 107. Binary Tree Level Order Traversal II 二叉树层次遍历反转

107. 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,null,null,15,7]

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

[
[15,7],
[9,20],
[3]
]

解题思路：
此题与Binary Tree Level Order Traversal相似，只是最后的结果有一个反转。

参考 http://qiaopeng688.blog.51cto.com/3572484/1834819 代码如下：

/**
* 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>> result;
queue<TreeNode *> current,next;
vector<int> level;
if(NULL == root)
return result;
current.push(root);

while(current.size())
{
while(current.size())
{
TreeNode *p;
p = current.front();
current.pop();
level.push_back(p->val);
if(p->left)
next.push(p->left);
if(p->right)
next.push(p->right);
}
result.push_back(level);
level.clear();
swap(current,next);
}
reverse(result.begin(),result.end());
//相对与Binary Tree Level Order Traversal只加了这一句。reverse()，反转
return result;
}
};