LeetCode103 Binary Tree Zigzag Level Order Traversal

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).（Medium）

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

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

分析：

层序遍历还要交叉输出。所以先采用队列进行层序编译（存放一层在队列中，每次循环先读这一队列的长度，然后以此把他们的值存在vector里，并且把存在的左右孩子压入队列）。

因为要交叉输出，所以采用一个flag，每次循环后flag正负号改变。负号时倒序输出。

代码：

 /**
  * 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>> result;
         if (root == nullptr) {
             return result;
         }
         queue<TreeNode* >que;
         que.push(root);
         int flag = ;
         while(!que.empty()) {
             int sz = que.size();
             vector<int> temp;
             for (int i = ; i < sz; ++i) {
                 TreeNode* cur = que.front();
                 que.pop();
                 temp.push_back(cur -> val);
                 if (cur -> left != nullptr) {
                     que.push(cur -> left);
                 }
                 if (cur -> right != nullptr) {
                     que.push(cur -> right);
                 }
             }
             if (flag == -) {
                  reverse(temp.begin(), temp.end());
                  result.push_back(temp);
                  flag = ;
             }
             else {
                 result.push_back(temp);
                 flag = -;
             }

         }
         return result;
     }
 };