1 """
2 Given a binary tree, determine if it is a valid binary search tree (BST).
3 Assume a BST is defined as follows:
4 The left subtree of a node contains only nodes with keys less than the node's key.
5 The right subtree of a node contains only nodes with keys greater than the node's key.
6 Both the left and right subtrees must also be binary search trees.
7 Example 1:
8 2
9 / \
10 1 3
11 Input: [2,1,3]
12 Output: true
13 Example 2:
14 5
15 / \
16 1 4
17 / \
18 3 6
19 Input: [5,1,4,null,null,3,6]
20 Output: false
21 Explanation: The root node's value is 5 but its right child's value is 4.
22 """
23 """
24 好题，提供四种解法,可与leetcode1305结合起来看https://www.cnblogs.com/yawenw/p/12284717.html
25 解法一：自己AC 中序遍历(递归） + sorted() + set()
26 思路是将二叉搜索树中序遍历存入nums
27 先set()处理，将输入为[1, 1]的情况排除
28 再sorted()与nums比较是否相等
29 因为如果为二叉搜索树 nums应该是没有重复值并且递增的
30 """
31 class TreeNode:
32 def __init__(self, x):
33 self.val = x
34 self.left = None
35 self.right = None
36
37 class Solution:
38 def isValidBST(self, root: TreeNode) -> bool:
39 if not root:
40 return True # bug input为[]时
41 res = self._inorder(root, [])
42 return res == sorted(set(res)) # bug 加上set是为了判断input[1,1]
43
44 def _inorder(self, root, nums):
45 if root.left:
46 self._inorder(root.left, nums)
47 nums.append(root.val)
48 if root.right:
49 self._inorder(root.right, nums)
50 return nums
51 """
52 解法二：递归
53 关键点在于初始化的时候
54 用无穷大和无穷小作为根结点的上下值
55 """
56 class Solution2:
57 def isValidBST(self, root):
58
59 return self._isValidBST(root)
60
61 def _isValidBST(self, root, low=float('-inf'), high=float('inf')): # ！！！关键的初始化
62 if not root:
63 return True
64 value = root.val
65 if not low < value < high:
66 return False
67 return self._isValidBST(root.left, low, value) and \
68 self._isValidBST(root.right, value, high)
69 """
70 解法三：将解法一的递归中序遍历变为栈
71 """
72 class Solution3:
73 def isValidBST(self, root):
74 stack = []
75 inorder = []
76 if not root:
77 return True
78 while stack or root: #非递归中序遍历
79 while root:
80 stack.append(root)
81 root = root.left
82 root = stack.pop() #bug这里写了 x = stack.pop(),影响上面while循环
83 inorder.append(root.val)
84 root = root.right #bug 这里写成了stack.append(root.right),原因没理解
85 return inorder == sorted(set(inorder))
86 """
87 解法四：将解法二的递归变为迭代(queue)
88 """
89 class Solution4:
90 def isValidBST(self, root):
91 if not root:
92 return True
93 queue = [(root, float('-inf'), float('inf'))]
94 while queue:
95 root, low, high = queue.pop(0)
96 value = root.val
97 if not low < value < high:
98 return False
99 if root.left: #if判别不能少
100 queue.append((root.left, low, value))
101 if root.right:
102 queue.append((root.right, value, high))
103 return True
