107. Binary Tree Level Order Traversal II

https://leetcode.com/problems/binary-tree-level-order-traversal-ii/description/
Solution 1. Recursive level traversal. Need to add a {} to res to increase res.size() when starting a new level, otherwise res[level] causes index error.
    void levelOrder(TreeNode* node, vector<vector<int>>& res, int level) {
        if(!node) return;
        if(level == res.size()) res.push_back({});
        res[level].push_back(node->val);
        levelOrder(node->left, res, level+1);
        levelOrder(node->right, res, level+1);
    }
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        levelOrder(root, res, 0);
        reverse(res.begin(), res.end());
        return res;
    }
Solution 2. Push each level to a queue. Use nullptr as a separator between levels.
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        queue<TreeNode*> q;
        TreeNode* node;
        q.push(root);
        q.push(nullptr);
        while(!q.empty()) {
            vector<int> level;
            while(node = q.front()){
                q.pop();
                level.push_back(node->val);
                if(node->left) q.push(node->left);
                if(node->right) q.push(node->right);
            }
            q.pop();
            q.push(nullptr);
            if(level.empty()) break;
            res.insert(res.begin(),level);
        }
        return res;
    }