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 = 1;
        while(!que.empty()) {
            int sz = que.size();
            vector<int> temp;
            for (int i = 0; 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 == -1) {
                 reverse(temp.begin(), temp.end());
                 result.push_back(temp);  
                 flag = 1;
            }
            else {
                result.push_back(temp);  
                flag = -1;
            }

        }
        return result;
    }
};