# LeetCode: 98. Validate Binary Search Tree

## 题目¶

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

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

Example 2:

```    5
/ \
1   4
/ \
3   6

Input: [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.
```

## 解法¶

### 中序遍历生成数组，判断数组是否有序¶

```def inorder(root):
if root is None:
return
inorder(root.left)

# print(root)

inorder(root.right)
```

```# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
def isValidBST(self, root):
self.node_vals = []
self.inorder(root)

pre = None
for v in self.node_vals:
if pre is not None and v <= pre:
return False
pre = v

return True

def inorder(self, root):
if root is None:
return
self.inorder(root.left)

self.node_vals.append(root.val)

self.inorder(root.right)
```

### 中序遍历直接判断是否是有效的 BST¶

```# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
def isValidBST(self, root):
self.pre = None
return self.inorder(root)

def inorder(self, root):
if root is None:
return True

if not self.inorder(root.left):
return False

if self.pre is not None and root.val <= self.pre.val:
return False

self.pre = root

return self.inorder(root.right)
```