** Notes: 这道题和 PathSum ii 比较类似，解法只需要改动一两行即可。 **Add to List

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Total Accepted: 101944 Total Submissions: 276629 Difficulty: Easy Contributor: LeetCode

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”]

**Credits:**Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res = new ArrayList<>();
        String cur = "";
        dfs(root, res, cur);
        return res;
    }
    
    private void dfs(TreeNode root, List<String> res, String cur) {
        if (root == null) {
            return ;
        }
        
        if (cur == "") {
          cur = cur + String.valueOf(root.val);
        } else {
          cur = cur + "->" + String.valueOf(root.val);    
        }
        
        if (root.left == null && root.right == null) {
            res.add(cur);
        }
        dfs(root.left, res, cur);
        dfs(root.right, res, cur);
        cur = cur.substring(0, cur.length() -1);
    }