leetcode129 Sum Root to Leaf Numbers 

 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

 

posted @ 2020-02-13 22:13  yawenw  阅读(135)  评论(0)    收藏  举报