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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

 

return its zigzag level order traversal as:

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

冥思苦想是如何打印？

结果发现是输出二维vector

事情变得容易很多

先写一个广序遍历二叉树

然后往里面填充

结果语法报错？？

发现写的有点蠢

每行结束多加一个NULL就可以解决

最后的最后按着一维去反转

/**
 * 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) {
        if (!root) 
            return {};
        vector<vector<int> > ans;
        vector<int>tmp;
        queue<TreeNode*>q;
        q.push(root);
        q.push(NULL);
        while(!q.empty()) {
            TreeNode* thi = q.front();
            q.pop();
            if(thi == NULL) {
                ans.push_back(tmp);
                tmp.resize(0);
                if(!q.empty()) {
                    q.push(NULL);
                }
            } else {
                tmp.push_back(thi->val);
                if(thi->left!=NULL) {
                    q.push(thi->left);
                }
                if(thi->right!=NULL) {
                    q.push(thi->right);
                }
            }

        }
        for(int i = 1;i < ans.size();i+=2) {
            reverse(ans[i].begin(),ans[i].end());
        }
        return ans;
    }
};