98.Validate Binary Search Tree:修订间差异
创建页面,内容为“=Description= {{LeetCode |id=validate-binary-search-tree |no=98 |difficulty=Medium |category=Tree |collection=MS |title=Validate Binary Search Tree |summary=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 s…” |
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Explanation: The root node's value is 5 but its right child's value is 4. | Explanation: The root node's value is 5 but its right child's value is 4. | ||
= Recursive = | |||
<syntaxhighlight lang="java"> | |||
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); | |||
} | |||
} | |||
</syntaxhighlight> | |||
[[Category:Algorithm]] | [[Category:Algorithm]] | ||
[[Category:LeetCode]] | [[Category:LeetCode]] |
2024年1月17日 (三) 17:18的版本
Description
#98 | Validate Binary Search Tree | Medium | ||
---|---|---|---|---|
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
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);
}
}