Lintcode 二叉树的锯齿形层次遍历

给出一棵二叉树，返回其节点值从底向上的层次序遍历（按从叶节点所在层到根节点所在的层遍历，然后逐层从左往右遍历）

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样例
给出一棵二叉树 {3,9,20,#,#,15,7},

3
/ \
9 20
/ \
15 7
按照从下往上的层次遍历为：

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

思路：在题 二叉树的层次遍历 http://blog.csdn.net/thinkerleo1997/article/details/77370131 的基础上引入了 一个脉冲变化的‘锯齿数’，当二叉树换层数后‘锯齿数’会取相反数，这时会改变这层二叉树读入的方式

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /*     * @param root: A Tree     * @return: A list of lists of integer include the zigzag level order traversal of its nodes' values.     */    vector<vector<int>> zigzagLevelOrder(TreeNode * root) {        // write your code here                vector< vector<int> > re;        if(root == NULL){            return re;        }        queue<TreeNode*> que;         que.push(root);         int should_len = 1;         int null_len = 0;        int sawtooth_num = 1; //引入 锯齿数        vector<int> now_s;         while(!que.empty()){            TreeNode *t = que.front();            que.pop();            if (t == NULL){                null_len ++;            }            else{                if(sawtooth_num == 1) //根据锯齿数来控制插入方式                {                    now_s.insert(now_s.end(), t->val);                            }                else                {                    now_s.insert(now_s.begin(), t->val);                }                que.push(t->left);                que.push(t->right);            }            if(should_len == null_len + now_s.size()                     && now_s.size() != 0){                re.insert(re.end(), now_s);                  now_s.clear();                  should_len *= 2;                  null_len *= 2;                  sawtooth_num = -sawtooth_num; //二叉树换层后锯齿数取负数                }            }        return re;    }};