【力扣hot100题】（042）验证二叉搜索树

方法是设置最大最小值，每次递归比较当前节点和这些值。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool judge(TreeNode* node,long int lower,long int upper){
        if(node==nullptr) return 1;
        if(node->val>=upper||node->val<=lower) return 0;
        return judge(node->left,lower,node->val)&&judge(node->right,node->val,upper);
    }
    bool isValidBST(TreeNode* root) {
        return judge(root,-2147483649,2147483648);
    }
};

其实还可以用中序遍历，也是递归。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
long int pre=-2147483649;
bool result=1;
    bool isValidBST(TreeNode* root) {
        if(!root) return 1;
        result=result&&isValidBST(root->left);
        if(pre>=root->val) result=0;
        pre=root->val;
        if(result==0) return 0;
        result=result&&isValidBST(root->right);
        return result;
    }
};