LeetCode98:Validate Binary Search Tree
阿新 • • 發佈:2019-01-05
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.
LeetCode:連結
題目要求去validate一個tree是不是Binary Search Tree,也就是所有left subtree的值都比現在node的值小,所有right subtree的值都比現在node的值大。
要注意的坑就是有可能是形如:
這就不是有效的binary search tree了,因為3雖然比8小,但是必須要比5大,所以在寫程式碼的時候不能只比較當前父節點的大小,還要考慮之前所有的父節點情況。所以我們必須要有兩個變數,lowerbound和upperbound。每次遞迴的時候更新這兩個變數。
# 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 """ lowerbound = float('-inf') upperbound = float('inf') return self.ValidBST(root, lowerbound, upperbound) def ValidBST(self, root, lowerbound, upperbound): if not root: return True if root.val <= lowerbound or root.val >= upperbound: return False return self.ValidBST(root.left, lowerbound, root.val) and self.ValidBST(root.right, root.val, upperbound)