Binary Tree Level Order Traversal

keyword: Queue, Swap, Vector< vector<int> >

// Binary Tree Level Order Traversal
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> > result;
        queue<TreeNode*> current, next;

        if(root == nullptr) {
            return result;
        } else {
            current.push(root);
        }

        while (!current.empty()) {
            vector<int> level; // elments in one level
            while (!current.empty()) {
                TreeNode* node = current.front();
                current.pop();
                level.push_back(node->val);
                if (node->left != nullptr) next.push(node->left);
                if (node->right != nullptr) next.push(node->right);
            }
            result.push_back(level);
            swap(next, current); //!!!!!!!!!!!!!
        }
        return result;
    }
};

// Binary Tree Level Order Traversal
// 递归版，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int>> result;
        traverse(root, 1, result);
        return result;
    }

    void traverse(TreeNode *root, size_t level, vector<vector<int>> &result) {
        if (!root) return;

        if (level > result.size())
            result.push_back(vector<int>());

        result[level-1].push_back(root->val);
        traverse(root->left, level+1, result);
        traverse(root->right, level+1, result);
    }
};