[Leetcode] 0098. Validate Binary Search Tree

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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:

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, 10 ⁴ ].
-2 ³¹ <= Node.val <= 2 ³¹ - 1

Table of contents

Python

				
					# time complexixty: O(n) # space complexity: O(n) import math from typing import Optional 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 dfs(node: Optional[TreeNode], low=-math.inf, high=math.inf): if node is None: return True if node.val <= low: return False if node.val >= high: return False return dfs(node.left, low, node.val) and dfs(node.right, node.val, high) return dfs(root) class Solution: def isValidBST(self, root: Optional[TreeNode]) -> bool: prev = [-math.inf] def dfs(node: Optional[TreeNode], prev): if not node: return True if not dfs(node.left, prev): return False if node.val <= prev[0]: return False prev[0] = node.val return dfs(node.right, prev) return dfs(root, prev) root1 = TreeNode(2) root1.left = TreeNode(1) root1.right = TreeNode(3) print(Solution().isValidBST(root1)) root2 = TreeNode(5) root2.left = TreeNode(1) root2.right = TreeNode(4) root2.right.left = TreeNode(3) root2.right.right = TreeNode(6) print(Solution().isValidBST(root2))

Python

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