[LeetCode]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],
]

... 感觉很囧在，前一题的基础上 加了一句 reverse（ret.begin(),ret.end()）

  .好像编程之美上有这题，没注意看。

    如果是每层的也要逆序的话，就把 left 和right  入队或入栈的顺序调换一下就行了

/**
 * 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> > ret;
			ret.clear();
			if(root == NULL)
				return ret;
			queue<TreeNode*> S;
			S.push(root);
			S.push(NULL);
			vector<int> tmp;
			while(!S.empty()){
				//travesal current level
				TreeNode* p = S.front();
				S.pop();
				if(p!=NULL)
				{
					tmp.push_back(p->val);
					if(p->left) 
						S.push(p->left);
					if(p->right)
						S.push(p->right);
				}else{
					if(!tmp.empty())
					{	
						S.push(NULL);
						ret.push_back(tmp);
						tmp.clear();
					}
				}
			}
			reverse(ret.begin(),ret.end());
            return ret;
    
    }
};