1 """
2 Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
3 An example is the root-to-leaf path 1->2->3 which represents the number 123.
4 Find the total sum of all root-to-leaf numbers.
5 Note: A leaf is a node with no children.
6 Example:
7 Input: [1,2,3]
8 1
9 / \
10 2 3
11 Output: 25
12 Explanation:
13 The root-to-leaf path 1->2 represents the number 12.
14 The root-to-leaf path 1->3 represents the number 13.
15 Therefore, sum = 12 + 13 = 25.
16 Example 2:
17 Input: [4,9,0,5,1]
18 4
19 / \
20 9 0
21 / \
22 5 1
23 Output: 1026
24 Explanation:
25 The root-to-leaf path 4->9->5 represents the number 495.
26 The root-to-leaf path 4->9->1 represents the number 491.
27 The root-to-leaf path 4->0 represents the number 40.
28 Therefore, sum = 495 + 491 + 40 = 1026.
29 """
30 """
31 本题两种解法
32 第一种为递归
33 """
34
35 class TreeNode:
36 def __init__(self, x):
37 self.val = x
38 self.left = None
39 self.right = None
40
41 class Solution1:
42 def sumNumbers(self, root):
43 self.res = 0 # ！！！res申请为全局变量,保存总的结果
44 self.sum(root, 0) # ！！！第一次传0值是为了递归的时候传入正确的根结点值
45 return self.res
46
47 def sum(self, root, value):
48 if root:
49 if root.left == None and root.right == None: # ！！！判断叶子结点
50 self.res += 10 * value + root.val
51 self.sum(root.left, 10 * value + root.val)
52 self.sum(root.right, 10 * value + root.val)
53
54 """
55 非递归的写法，先用tuple存储(value, root)
56 然后入队列遍历
57 """
58
59 class Solution2:
60 def sumNumbers(self, root):
61 res = 0
62 if root == None:
63 return res
64 queue = []
65 x = (0, root) # tuple元组,(父亲结点的值,结点)
66 queue.append(x)
67 while queue:
68 value, node = queue.pop() # 头结点出队
69 if node:
70 if not node.left and not node.right: # 判断叶子结点，加各个路径的和
71 res = res + node.val + value * 10
72 queue += [(node.val + 10 * value, node.left), (node.val + 10 * value, node.right)]
73 # 层次遍历 也可以用queue.extend([(, ), (, )])来写
74 return res
