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.
A single node tree is a BST
Example
An example:
2
/ \
1 4
/ \
3 5
The above binary tree is serialized as {2,1,4,#,#,3,5} (in level order).
Solution
常犯错误就是忽略的等于号
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: The root of binary tree.
@return: True if the binary tree is BST, or false
"""
def isValidBST(self, root):
isValid,max,min=self.isValidBSTHelper(root)
return isValid
def isValidBSTHelper(self, root):
# write your code here
if root is None:
return True, -sys.maxsize, sys.maxsize
left = self.isValidBSTHelper(root.left)
right = self.isValidBSTHelper(root.right)
if left[0] == False or right[0] == False:
return False, 0, 0
if root.left and root.val <= left[1]:
return False, 0, 0
if root.right and root.val >= right[2]:
return False, 0, 0
return True, max(root.val,right[1]), min(root.val, left[2])