LeetCode098. Validate Binary Search Tree

最新推荐文章于 2020-10-15 09:59:45 发布

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本文探讨了如何判断一个二叉树是否为有效的二叉搜索树（BST）。通过定义BST的特性，即左子树节点小于根节点，右子树节点大于根节点，且左右子树也必须是BST，提出了一种解决方案。利用中序遍历二叉树的特性，若遍历结果为升序且无重复值，则该二叉树为有效的BST。

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.

二叉搜索树按照树的中序遍历应该是升序排列的.

所以用中序遍历后排序与原结果进行比对.还要注意不能有重复数字出现.

class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root is None:return True
        res = []
        result = self.helper(root, res)

        return sorted(result)==result and len(result)==len(set(result))
        
    def helper(self, root, res):
        if root is None:return root
        if root:
            self.helper(root.left, res)
            res.append(root.val)
            self.helper(root.right, res)
        return res