98.Validate Binary Search Tree:修订间差异

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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);
    }
}