# Sum Root to Leaf Numbers(recrusive)

Given a binary tree containing digits from`0-9`only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path`1->2->3`which represents the number`123`.

Find the total sum of all root-to-leaf numbers.

**Note:** A leaf is a node with no children.

**Example:**

```
Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
```

**Example 2:**

```
Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
```

分析

叶子只负责算Pathsum。root只负责把左右相加。值得注意的做法

```
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sumNumbers(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        return self.dfs(root,0)

    def dfs(self,root,pathsum):
        if not root:
            return 0
        #leaf get pathsum
        if not root.left and not root.right:
            return root.val+pathsum*10
        #root only calculate
        return self.dfs(root.left, pathsum*10+root.val)+self.dfs(root.right, pathsum*10+root.val)
```


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