[Leetcode]Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal My Submissions Question
Total Accepted: 47731 Total Submissions: 176275 Difficulty: Medium
Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

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依旧是对树进行层序遍历，之后对序号为奇数的向量进行一次翻转。
不出意料的一次AC，自从看完《剑指offer》之后，觉得学会了边写代码边检查，各种边界条件测试在脑海里一遍遍过。

/**
 * 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>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int> >    res;
        if(!root)   return res;
        queue<TreeNode*>    q;
        q.push(root);
        vector<int> vi;
        while(!q.empty()){
            int size = q.size();
            for(int i = size;i > 0;--i){
                TreeNode* t = q.front();
                q.pop();
                vi.push_back(t->val);
                if(t->left) q.push(t->left);
                if(t->right)    q.push(t->right);
            }
            res.push_back(vi);
            vi.clear();
        }
        for(int i = 0;i < res.size();++i){
            if(i % 2){
                reverse(res[i].begin(),res[i].end());
            }
        }
        return res;
    }
};