# 98. Validate Binary Search Tree

#### Medium

***

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:**

![](https://assets.leetcode.com/uploads/2020/12/01/tree1.jpg)

```
Input: root = [2,1,3]
Output: true
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/12/01/tree2.jpg)

```
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`

#### Solution 1 : Without Any Space

```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]) -> bool:
        def recursion(root, floor = -float("inf"), ceil = float("inf")):
            if root is None:
                return True
            if root.val <= floor or root.val >= ceil:
                return False
            return recursion(root.left, floor , root.val) and recursion(root.right, root.val, ceil)
        return recursion(root)
```

```python
class Solution:
    def isValidBST(self, root: Optional[TreeNode], floor = -float("inf"), ceil = float("inf")) -> bool:
        if root is None:
                return True
        if root.val <= floor or root.val >= ceil:
            return False
        return self.isValidBST(root.left, floor , root.val) and self.isValidBST(root.right, root.val, ceil)
```

#### Solution 2 : With Space

```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]) -> bool:
        arr = []
        self.inorder(root, arr)
        for index in range(1, len(arr)):
            if arr[index] <= arr[index-1]:
                return False
        return True
        
    def inorder(self, root, arr):
        if not root:
            return
        self.inorder(root.left, arr)
        arr.append(root.val)
        self.inorder(root.right, arr)
```