-
Notifications
You must be signed in to change notification settings - Fork 27
/
Copy path98.ValidateBinarySearchTree.py
48 lines (41 loc) · 1.47 KB
/
98.ValidateBinarySearchTree.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
"""
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
- The left subtree of a node contains only nodes with keys less than
the node's key.
- The right subtree of a node contains only nodes with keys greater
than the node's key.
- Both the left and right subtrees must also be binary search trees.
Example:
5
/ \
1 4
/ \
3 6
Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.
"""
#Difficulty: Medium
#75 / 75 test cases passed.
#Runtime: 44 ms
#Memory Usage: 17 MB
#Runtime: 44 ms, faster than 77.80% of Python3 online submissions for Validate Binary Search Tree.
#Memory Usage: 17 MB, less than 5.06% of Python3 online submissions for Validate Binary Search Tree.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isValidBST(self, root: TreeNode) -> bool:
order = []
self.validator(root, order)
return len(order) == len(set(order)) and order == sorted(order)
def validator(self, root, order):
if not root:
return 0
self.validator(root.left, order)
order.append(root.val)
self.validator(root.right, order)