Medium
Given the root
of a binary tree, determine if it is a valid binary search tree (BST).
A valid BST is defined as follows:
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.
Constraints:
[1, 104]
.-231 <= Node.val <= 231 - 1
#include <climits>
/**
* 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 isValidBST(TreeNode* root) {
return solve(root, LONG_MIN, LONG_MAX);
}
private:
bool solve(TreeNode* root, long left, long right) {
if (root == nullptr) {
return true;
}
if (root->val <= left || root->val >= right) {
return false;
}
return solve(root->left, left, root->val) && solve(root->right, root->val, right);
}
};