98. Validate Binary Search Tree
Category: /leetcodeGiven 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.
Solution:
- preorder recursion
def isValidBST(self, root: TreeNode) -> bool:
def is_valid(root, low, high):
if not root: return True
if root.val <= low or root.val >= high:
return False
return is_valid(root.left, low, root.val) \
and is_valid(root.right, root.val, high)
return is_valid(root, float("-inf"), float("inf"))
- follow up: if asked to write iteration version
def isValidBST(self, root: TreeNode) -> bool:
if root is None: return True
stack = [(root, float("-inf"), float("inf")), ]
while stack:
root, lower, upper = stack.pop()
if not root:
continue
val = root.val
if val <= lower or val >= upper:
return False
stack.append((root.left, lower, val))
stack.append((root.right, val, upper))
return True
inorder is an idea of sorting traversal.
def isValidBST(self, root: TreeNode) -> bool:
stack, inorder = [], float("-inf")
while stack or root:
while root:
stack.append(root)
root = root.left
root = stack.pop()
# if next element in inorder traversal is smaller
# then it is not BST
if root.val <= inorder: return False
inorder = root.val
root = root.right
return True
