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.
二叉搜索树也是二叉排序树,就是树的根结点值大于左子结点值,小于右子结点值。
那么先判断根结点是否满足条件,然后递归遍历左右子结点。
# 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: TreeNode) -> bool:
return self.isBST(root,None,None)
def isBST(self,root,l,r):
if not root:
return True
if l and l.val >= root.val:
return False
if r and r.val <= root.val:
return False
return self.isBST(root.left,l,root) and self.isBST(root.right,root,r)