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).

 def isValidBST(self, root):
         """
         :type root: TreeNode
         :rtype: bool
         """
         maxInt = 2147483647
         minInt = -2147483648

         return self.isBSTHelper(root, minInt, maxInt)

     def isBSTHelper(self, root, minVal, maxVal):
         if not root:
             return True

         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):
             return True
         else:
             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。

 class Solution(object):
     def isValidBST(self, root):
         """
         :type root: TreeNode
         :rtype: bool
         """
         ans = []
         self.inOrder(root, ans)

         n = len(ans)
         if n <= 1:
             return True
         for i in range(1,n):
             if ans[i] <= ans[i-1]:
                 return False
         return True

     def inOrder(self, root, ans):
         if not root:
             return ans
         else:
             self.inOrder(root.left, ans)
             ans.append(root.val)
             self.inOrder(root.right, ans)