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)