[leetcode] Validate Binary Search Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/mungco/article/details/10070097

本文介绍了一种通过递归方法验证二叉树是否为有效的二叉搜索树的方法。文章详细阐述了如何检查二叉树节点的左子树和右子树是否满足二叉搜索树的定义，并提供了具体的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

class Solution {
public:
    bool isValidBST(TreeNode *root) {
        // Start typing your C/C++ solution below
		// DO NOT write int main() function
		if(root==NULL)
			return true;
        TreeNode *lm,*rm;
        lm=leftmost(root);
        rm=rightmost(root);
		if(root->left!=NULL && root->left->val>=root->val || lm && lm->val >= root->val)
			return false;
		if(root->right!=NULL && root->right->val<=root->val || rm && rm->val<=root->val)
			return false;
		return isValidBST(root->left)&&isValidBST(root->right);
	}
    TreeNode* rightmost(TreeNode *root){
        if(root->right)
            root=root->right;
        else
            return NULL;
        while(root->left)
            root=root->left;
        return root;
    }
    TreeNode* leftmost(TreeNode *root){
        if(root->left)
            root=root->left;
        else
            return NULL;
        while(root->right)
            root=root->right;
        return root;
    }
};