98. Validate Binary Search Tree - Medium
阿新 • • 發佈:2018-12-31
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.
recursion
給每個節點定義一個區間,如果節點的值落在區間外,返回false,否則繼續遞迴左右節點,區間上下限相應調整
注意!!min, max如果用int會溢位,要用long
time: O(n), space: O(n)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }*/ class Solution { public boolean isValidBST(TreeNode root) { return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE); } public boolean isValid(TreeNode node, long min, long max) { if(node == null) { return true; } if(node.val <= min || node.val >= max) { return false; } return isValid(node.left, min, node.val) && isValid(node.right, node.val, max); } }