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 path1->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->2represents the number 12.
The root-to-leaf path1->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.
思路
这道题我看到之后第一想到的就是使用递归进行解决，递归的结束条件就是当遍历到左右节点都为空时，直接返回结果。递归条件就是对左右子节点进行递归。
解决代码
1 # Definition for a binary tree node.
2 # class TreeNode(object):
3 # def __init__(self, x):
4 # self.val = x
5 # self.left = None
6 # self.right = None
7
8 class Solution(object):
9 def sumNumbers(self, root):
10 """
11 :type root: TreeNode
12 :rtype: int
13 """
14 if not root: # 根节点为空直接返回
15 return 0
16 res = self.GetResult(root, 0)
17 return res
18
19
20 def GetResult(self, root, num):
21 if not root.left and not root.right: # 左右节点为空返回当前结果。
22 return num+ root.val
23 left = right = 0 # 左右子树的值
24 if root.left:
25 left = self.GetResult(root.left, (num+root.val)*10)
26 if root.right:
27 right = self.GetResult(root.right, (num+root.val)*10)
28 return left+right # 左右子树结果相加