a.cpp

/**
 * 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;
       
        queue <TreeNode*> q;
       
        if(root == NULL)
            return {};
        if(root->left == NULL && root->right == NULL)
            return {{root->val}};
       
        int level = 0;
        q.push(root);
        while(!q.empty()) {
           
            int size = q.size();
           
            result.push_back(vector<int>());
                             
            for(int i = 0; i < size; ++i) {
   
                TreeNode *node = q.front();
                q.pop();
                result[level].push_back(node->val);
               
                if(node->left != NULL)
                    q.push(node->left);
                if(node->right != NULL)
                    q.push(node->right);
                   
                 }
           
            ++level;
       
      }
       
       
       
        for(int i = 0; i < result.size(); ++i) {
            if(i % 2 == 1) {
                reverse(result[i].begin(), result[i].end());
            }
        }
       
       
     
        return result;
       
    }
};

line 20: unnecessary exit? you can do that in while loop
line 29: you know the size of the vector. You should use the constructor where you can specify the size for optimisation.
line 44: level is not used anywhere
line 50: Can you do it without using this extra for loop and reverse statements? Takes additional O(n) for no reason.