Problem
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 1:
2
/ \
1 3
Input: [2,1,3]
Output: true
Example 2:
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.
Solution
Important:
You don’t only valid the current node with its left and right value. You have to:
- pass the new high bound for left branch, because all the values in the left branch must be less than this high bound, low bound remains
- pass the new low bound for right branch, because all the values in the right branch must be greater than this low bound, high bound remains
class Solution:
def isValidBST(self, root):
return self.isValidBSTBound(root, -sys.maxsize, sys.maxisze)
def isValidBSTBound(self, root, lb, hb):
if not root:
return True
if lb < root.val < hb:
return self.isValidBSTBound(root.left, lb, root.val) and self.isValidBSTBound(root.right, root.val, hb)
else:
return False