98.Validate Binary Search Tree:修订间差异
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{{Submission|runtime=0ms|memory=44.65MB|rp=100|mp=9.94}} | |||
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2024年1月17日 (三) 17:19的最新版本
Description
#98 | Validate Binary Search Tree | Medium | ||
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Tree | MS | |||
Given the root of a binary tree, determine if it is a valid binary search tree (BST). |
Given the root of a binary tree, determine if it is a valid binary search tree (BST).
A valid 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: root = [2,1,3]
Output: true
Example 2:
Input: root = [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.
Recursive
Runtime 0ms |
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Memory 44.65MB |
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class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, null, null);
}
private boolean isValidBST(TreeNode root, Integer lessThan, Integer greaterThan) {
if(root == null) return true;
if(lessThan != null && root.val >= lessThan) return false;
if(greaterThan != null && root.val <= greaterThan) return false;
if(root.left != null && root.left.val >= root.val) return false;
if(root.right != null && root.right.val <= root.val) return false;
return isValidBST(root.left, root.val, greaterThan) && isValidBST(root.right, lessThan, root.val);
}
}