98. Validate Binary Search Tree
Medium 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.
Example 1
Input: root = [2,1,3]
Output: true
Example 2
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
How to solve the problem
Code
- Recursion
Python
# 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],left=-inf, right=inf) -> bool:
if root is None:
return True
x = root.val
return left < x < right and self.isValidBST(root.left, left, x) and self.isValidBST(root.right, x, right)
Python
# 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], left=-inf, right=inf) -> bool:
if not root:
return True
if not (left < root.val < right):
return False
return self.isValidBST(root.left, left, root.val) and self.isValidBST(root.right, root.val, right)
Complexity
- Time complexity: O(n), n == number of nodes
- Space complexity: O(n)
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