1 """
2 Given two binary search trees root1 and root2.
3 Return a list containing all the integers from both trees sorted in ascending order.
4 Example 1:
5 Input: root1 = [2,1,4], root2 = [1,0,3]
6 Output: [0,1,1,2,3,4]
7 Example 2:
8 Input: root1 = [0,-10,10], root2 = [5,1,7,0,2]
9 Output: [-10,0,0,1,2,5,7,10]
10 Example 3:
11 Input: root1 = [], root2 = [5,1,7,0,2]
12 Output: [0,1,2,5,7]
13 Example 4:
14 Input: root1 = [0,-10,10], root2 = []
15 Output: [-10,0,10]
16 Example 5:
17 Input: root1 = [1,null,8], root2 = [8,1]
18 Output: [1,1,8,8]
19 """
20 class TreeNode:
21 def __init__(self, x):
22 self.val = x
23 self.left = None
24 self.right = None
25
26 """
27 此题提供三种思路，与leetcode98类似，https://www.cnblogs.com/yawenw/p/12376942.html
28 第一种是层次遍历，
29 然后对存入的值sorted排序
30 传送门:https://blog.csdn.net/qq_17550379/article/details/103838538
31 """
32 class Solution1:
33 def getAllElements(self, root1, root2):
34 q, res = [root1, root2], []
35 while q:
36 cur = q.pop(0)
37 if cur:
38 res.append(cur.val)
39 if cur.left != None:
40 q.append(cur.left)
41 if cur.right != None:
42 q.append(cur.right)
43 return sorted(res)
44
45 """
46 第二种是利用二叉搜索树的条件，
47 对两个树分别中序遍历。这样两个list分别有序
48 再进行归并排序
49 """
50 class Solution2:
51 def getAllElements(self, root1, root2):
52 q1, q2 = [], []
53 res = []
54 # 中序遍历
55 def inorder(root, q):
56 if root:
57 inorder(root.left, q)
58 q.append(root.val)
59 inorder(root.right, q)
60 inorder(root1, q1)
61 inorder(root2, q2)
62 # 归并排序的方法
63 while q1 or q2:
64 if not q1:
65 res += q2
66 break
67 if not q2:
68 res += q1
69 break
70 else:
71 res.append(q1.pop(0) if q1[0] < q2[0] else q2.pop(0))
72 return res
73
74 """
75 第三种是：
76 先中序遍历(代替层次遍历)
77 再sorted(代替归并排序)
78 """
79 class Solution3:
80 def getAllElements(self, root1, root2):
81 res = []
82 def inOrder(root):
83 if root: #!!!bug 没有写次if语句
84 inOrder(root.left)
85 res.append(root.val)
86 inOrder(root.right)
87 inOrder(root1)
88 inOrder(root2)
89 return sorted(res)
