leetcode 366 Find Leaves of Binary Tree

Conventional thinking is to take over dfs identify a leaf node, record, delete the leaf node, repeatedly until the tree empty, time complexity is o (n!)

However, we can record from the current node to the leaf node, to complete the subject of the request only once DFS, time complexity is o (n)

public List<List<Integer>> res;
public List<List<Integer>> findLeaves(TreeNode root){
    res = new ArrayList<>();
    if(root == null)
        return res;
    dfs(root);
    return res;
}
private int dfs(TreeNode root){
    if(root == null)
        return 0;
    int left = dfs(root.left);
    int right = dfs(root.right);
    int depth = Math.max(left, right) + 1;
    if(depth > res.size()){
        res.add(new ArrayList<>(){root.val});
    }else{
        res.get(depth-1).add(root.val);
    }
    return depth;
}