题目:
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:
Input: 2 / 1 3 Output: true
Example 2:
5 / 1 4 / 3 6 Output: false Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value is 5 but its right child's value is 4.
思路:
要判定一棵树是二叉搜索树,需要判定两点,一个是当前节点的值小于右子树的最小值,大于左子树的最大值;一个是其左右子树仍满足这个性质。
这与“判定一棵树是否是平衡二叉树”问题类似,于是想到一种自底向上的解法,依次判定左右子树是否为二叉搜素树,并向上传递左右子树的最小值、最大值及是否为二叉搜索树。代码如下:
class Solution(object): def isValidBST(self, root): """ :type root: TreeNode :rtype: bool """ if not root: return True _i, _j, res = self.checkValid(root) return res def checkValid(self, root): max_val, min_val = root.val, root.val if root.left: left_max, left_min, left_res = self.checkValid(root.left) if not left_res or left_max >= root.val: return 0, 0, False min_val = left_min if root.right: right_max, right_min, right_res = self.checkValid(root.right) if not right_res or right_min <= root.val: return 0, 0, False max_val = right_max return max_val, min_val, True
但这种传递显得比较乱,要写的代码比较长,因此使用另一个方式,自顶向下的判断,并将判断结果自底向上传递。代码如下,会简洁一些:
class Solution(object): def isValidBST(self, root): """ :type root: TreeNode :rtype: bool """ return self.checkValid(root, float('-inf'), float('+inf')) def checkValid(self, root, min_val, max_val): if not root: return True if root.val <= min_val or root.val >= max_val: return False return (self.checkValid(root.left, min_val, root.val) and self.checkValid(root.right, root.val, max_val))
另, float('+inf')表示正无穷的这种用法也比较有趣。