LeetCode 98. Validate Binary Search Tree
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原题
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 3Binary tree
[2,1,3]
, return true.
Example 2:
1 / \\ 2 3Binary tree
[1,2,3]
, return false.
解题思路
思路一:
- 先将二叉树进行中序遍历,接着判断遍历的数据是否为递增的即可(较简单)
思路二:
- 采用二叉搜索树的定义来解题(较难)
- 难点:每个节点都得跟两个数进行比较,它必须在这两个数的范围之内,其中一个是它的父节点,另外一个是它的某个祖先节点
- 这个树是从上往下遍历的,遍历到每个节点时,都要把它限定在一个已知的范围内,不管他下面的节点,只根据他上面的节点来找出这个范围
- 这里我采用继承的方法,首先定义根节点的范围为正负无穷大,节点每一个左节点的最大值为它的父节点,最小值继承它父节点的最小值(右节点相似),由此就可以构造出一个递归程序。
完整代码
思路一
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None # 先将二叉搜索树进行中序遍历,接着判断是否满足规则 class Solution(object): def isValidBST(self, root): """ :type root: TreeNode :rtype: bool """ self.ret = [] self.helper(root) for i in xrange(1, len(self.ret)): if self.ret[i] <= self.ret[i-1]: return False return True def helper(self, node): if node == None: return self.helper(node.left) self.ret.append(node.val) self.helper(node.right)
思路二
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def isValidBST(self, root): """ :type root: TreeNode :rtype: bool """ return self.helper(root, -float("inf"), float("inf")) def helper(self, node, low, high): if node == None: return True if node.val <= low or node.val >= high: return False return self.helper(node.left, low, node.val) and self.helper(node.right, node.val, high)
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