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70 lines
2.0 KiB
Plaintext
70 lines
2.0 KiB
Plaintext
---
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title: '98. Validate Binary Search Tree'
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description: Given the root of a binary tree, determine if it is a valid binary search tree (BST)
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icon: dot
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topics:
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- { name: "Tree", slug: "tree" }
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Binary Search Tree", slug: "binary-search-tree" }
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- { name: "Binary Tree", slug: "binary-tree" }
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video: "s6ATEkipzow"
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issue: "https://github.com/prdlk/leetcode/issues/111"
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sidebar:
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label: 'Validate Binary Search Tree'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `root = [2,1,3]`
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- Output: `true`
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### Example 2:
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- Input: `root = [5,1,4,null,null,3,6]`
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- Output: `false`
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- Explanation: The `root` node's value is `5` but its right child's value is `4`.
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### Constraints:
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- The number of nodes in the tree is in the range [1, 10^4].
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- `-2^31 <= Node.val <= 2^31 - 1`
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## Approach
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```mermaid
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flowchart TD
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S(["isValidBST(root)"]) --> C["valid(root, float('-inf'), float('inf')) — the root may hold any value"]
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C --> E(["return that result"])
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subgraph valid["valid(node, low, high)"]
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V1{"not node?"}
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V1 -- yes --> V2(["return True"])
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V1 -- no --> V3{"not (low < node.val < high)?"}
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V3 -- yes --> V4(["return False — node breaks a bound set by an ancestor"])
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V3 -- no --> V5["valid(node.left, low, node.val) and valid(node.right, node.val, high) — each child tightens one bound"]
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V5 --> V6(["return that result"])
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end
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C --> V1
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```
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## Solution
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```py
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def isValidBST(self, root: Optional[TreeNode]) -> bool:
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def valid(node, low, high):
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if not node:
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return True
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if not (low < node.val < high):
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return False
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return valid(node.left, low, node.val) and valid(node.right, node.val, high)
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return valid(root, float("-inf"), float("inf"))
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```
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