Given a digit string, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below.
Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Solution1：BFS

廣度優先遍歷，queue實現
Time Complexity: O(N) Space Complexity: O(n) worst緩存

Solution2：DFS with level

DFS過程中不同level的結果list 保存到 對應ArrayList<list> result[level]中
Time Complexity: O(N) Space Complexity: O(n) worst緩存

Solution1 Code：

class Solution {
    //BFS 最要在Node進入queue前就進行非空判斷，使得進入queue的node本身都非null
    public List<List<Integer>> levelOrder(TreeNode root) {
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> result = new LinkedList<List<Integer>>();
        
        if(root == null) return result;
        
        queue.offer(root);
        while(!queue.isEmpty()) {
            int num_on_level = queue.size();
            List<Integer> cur_result = new LinkedList<Integer>();
            for(int i = 0; i < num_on_level; i++) {
                TreeNode cur = queue.poll();
                if(cur.left != null) queue.add(cur.left);
                if(cur.right != null) queue.add(cur.right);
                cur_result.add(cur.val);
            }
            result.add(cur_result);
        }
        return result;
    }
}

Solution2 Code：

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        dfs(result, root, 0);
        return result;
    }
    
    public void dfs(List<List<Integer>> result, TreeNode root, int level) {
        if (root == null) return;
        if (level == result.size()) {
            result.add(new LinkedList<Integer>());
        }
        result.get(level).add(root.val);
        dfs(result, root.left, level+1);
        dfs(result, root.right, level+1);
    }
}