Given the root of a binary tree, determine if it is a valid binary search tree (BST).
A valid 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 1:
2
/ \
1 3
Input: root = [2,1,3]
Output: true
Example 2:
5
/ \
1 4
/ \
3 6
Input: root = [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.
Constraints:
- The number of nodes in the tree is in the range
[1, 104].
-231 <= Node.val <= 231 - 1
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# 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: Optional[TreeNode]) -> bool:
def validate(
node: Optional[TreeNode],
low=-math.inf,
high=math.inf
) -> bool:
if not node:
return True
if node.val <= low or node.val >= high:
return False
return (
validate(node.left, low, node.val) and
validate(node.right, node.val, high)
)
return validate(root)
'''
In order
'''
class Solution:
def isValidBST(self, root: TreeNode) -> bool:
def inorder(root):
if not root:
return True
if not inorder(root.left):
return False
if root.val <= self.prev:
return False
self.prev = root.val
return inorder(root.right)
self.prev = -math.inf
return inorder(root)
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