Leetcode 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

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

Example 2:

    1
   / \
  2   3

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

对每一个节点使用一个upper bound and lower bound. 时间复杂度 O(N).

 1 def isValidBST(self, root):
 2         """
 3         :type root: TreeNode
 4         :rtype: bool
 5         """
 6         maxInt = 2147483647
 7         minInt = -2147483648
 8         
 9         return self.isBSTHelper(root, minInt, maxInt)
10         
11     def isBSTHelper(self, root, minVal, maxVal):
12         if not root:
13             return True
14         
15         if root.val <= maxVal and root.val >= minVal and self.isBSTHelper(root.left, minVal, root.val - 1) and self.isBSTHelper(root.right, root.val + 1, maxVal):
16             return True
17         else:
18             return False

 思路二：如果中序遍历一下tree，BST应该给出一个递增序列。注意中序遍历的非递归算法。

 1 class Solution(object):
 2     def isValidBST(self, root):
 3         """
 4         :type root: TreeNode
 5         :rtype: bool
 6         """
 7         ans = []
 8         self.inOrder(root, ans)
 9         
10         n = len(ans)
11         if n <= 1:
12             return True
13         for i in range(1,n):
14             if ans[i] <= ans[i-1]:
15                 return False
16         return True
17                 
18     def inOrder(self, root, ans):
19         if not root:
20             return ans
21         else:
22             self.inOrder(root.left, ans)
23             ans.append(root.val)
24             self.inOrder(root.right, ans)    

 

转载于:https://www.cnblogs.com/lettuan/p/6196402.html