Sum Root to Leaf Numbers

 1 public class Solution {
 2     public int sumNumbers(TreeNode root) {
 3         if (root == null) {
 4             return 0;
 5         }
 6         return dfs(root, 0);
 7     }
 8 
 9     //dfs 函数 是求以base为基数的从当前节点到叶子节点的和。
10     private int dfs(TreeNode root, int base) {
11         if (root == null) {
12             return 0;
13         }
14         int sum = base * 10 + root.val;
15         if (root.left == null && root.right == null) {
16         return sum; 
17         }
18         return dfs(root.left, sum) + dfs(root.right, sum);
19     }
20 }