95题 Validate Binary Search Tree

最新推荐文章于 2025-12-05 17:02:52 发布

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该博客讨论了一种算法，用于检查给定的二叉树是否符合二叉搜索树的定义。二叉搜索树的特性是左子树上的节点值小于根节点，右子树上的节点值大于根节点。解决方案中，`isValidBST`方法通过递归地检查每个子树来确定整个树是否有效，并使用`helper`辅助函数来更新最小值和最大值。如果在递归过程中发现违反了二叉搜索树的规则，返回false；否则返回true。

Validate Binary Search Tree

Description
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.
A single node tree is a BST

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
class ResultType{
    boolean is_bst ;
    int minValue,maxValue ;
    public ResultType(boolean is_bst , int minValue , int maxValue){
        this.is_bst = is_bst ;
        this.minValue = minValue ;
        this.maxValue = maxValue ;

    }

}
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if the binary tree is BST, or false
     */
    public boolean isValidBST(TreeNode root) {
        // write your code here
        ResultType result = helper(root) ;
        return result.is_bst ;
    }
    public ResultType helper(TreeNode root){
        if(root == null){
            return new ResultType(true , Integer.MAX_VALUE ,Integer.MIN_VALUE );
        }
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        if(! left.is_bst || ! right.is_bst){
            return new ResultType(false , 0 ,0 );
        }
        if(root.left != null && left.maxValue >= root.val || root.right != null && right.minValue <= root.val){
            return new ResultType(false , 0 ,0 );
        }
        return new ResultType(true , Math.min(root.val , left.minValue) ,Math.max(root.val , right.maxValue) );
    }
}