# Leetcode: Validate Binary Search Tree

Basic tree datastructure

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
```

Binary tree [2,1,3], return true.

```Example 2:
1
/ \
2   3
```

Binary tree [1,2,3], return false.

Github: code.dennyzhang.com

Credits To: leetcode.com

```## Blog link: https://code.dennyzhang.com/validate-binary-search-tree
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
import sys
return self._isValidBST(root, -sys.maxsize-1, sys.maxsize)

def _isValidBST(self, root, min_value, max_value):
if root is None:
return True

if root.val <= min_value or root.val >= max_value:
return False

if root.left:
if self._isValidBST(root.left, min_value, root.val) is False:
return False

if root.right:
if self._isValidBST(root.right, root.val, max_value) is False:
return False

return True

def isValidBST_v1(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
## Idea: DFS recursive
## Complexity: Time O(n), Space O(log(n))
list_value = self.getBST(root)
for i in range(0, len(list_value)-1):
if list_value[i+1] <= list_value[i]:
return False
return True

def getBST(self, root):
if root is None:
return []

res = []
if root.left:
res += self.getBST(root.left)

res.append(root.val)

if root.right:
res += self.getBST(root.right)
return res
```

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