257. Binary Tree Paths

Problem:

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

["1->2->5", "1->3"]

Analysis:
Typical DFS, but to remember the ending condition is root.left == null and root.right == null.
Solution:
After some optimization of code structure;

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res =  new ArrayList<>();
        if (root == null) return res;
        dfs(root, new StringBuilder(), res);
        return res;
    }
    
    private void dfs(TreeNode root, StringBuilder sb, List<String> res) {
        if (root == null) return;
        int len = sb.length();
        sb.append(Integer.toString(root.val));
        if (root.left == null && root.right == null) {
            res.add(sb.toString());
        } else {
            sb.append("->");
            dfs(root.left, sb, res);
            dfs(root.right, sb, res);
        }
        sb.setLength(len);
    }
}