102. Binary Tree Level Order Traversal Solution using 2 queues

levelOrder2.cpp

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> even;
        queue<TreeNode*> odd;
        vector<vector<int>> result;
        
        even.push(root);
        
        while(!even.empty() || !odd.empty()){
            vector<int> level;
           
            if(odd.empty()){
                // while even not empty, transfer kids of falling out nodes to odd 
                while(!even.empty()){
                    TreeNode* curr = even.front(); even.pop();
                   
                    if(curr){
                        level.push_back(curr->val);
                        
                        odd.push(curr->left);
                        odd.push(curr->right);
                    }
                }
            }
            else if(even.empty()){
                // if even is empty, transfer kids of falling out nodes to even | keep juggling 
                while(!odd.empty()){
                    TreeNode* curr = odd.front(); odd.pop();
                   
                    if(curr){
                        level.push_back(curr->val);
                        
                        even.push(curr->left);
                        even.push(curr->right);
                    }
                }
            }
            
            result.push_back(level);
        }
        
        result.pop_back();
        return result;
    }
};