0
Software Development

## 【Day 30】#98 - Validate Binary Search Tree

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.
``````

## 解法一(Recursion)

``````class Solution:
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
def helper(node, lower = float('-inf'), upper = float('inf')):
if not node:
return True

val = node.val
if val <= lower or val >= upper:
return False

if not helper(node.right, val, upper):
return False
if not helper(node.left, lower, val):
return False
return True

return helper(root)
``````

## 解法二(Iteration)

``````class Solution:
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if not root:
return True

stack = [(root, float('-inf'), float('inf')), ]
while stack:
root, lower, upper = stack.pop()
if not root:
continue
val = root.val
if val <= lower or val >= upper:
return False
stack.append((root.right, val, upper))
stack.append((root.left, lower, val))
return True
``````

LeetCode刷題日記30