98. Validate Binary Search Tree

Recursion

/*
 * @lc app=leetcode id=98 lang=cpp
 *
 * [98] Validate Binary Search Tree
 *
 * https://leetcode.com/problems/validate-binary-search-tree/description/
 *
 * algorithms
 * Medium (33.60%)
 * Likes:    17453
 * Dislikes: 1402
 * Total Accepted:    2.8M
 * Total Submissions: 8.1M
 * Testcase Example:  '[2,1,3]'
 *
 * 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.
 *
 *
 *
 * Constraints:
 *
 *
 * The number of nodes in the tree is in the range [1, 10^4].
 * -2^31 <= Node.val <= 2^31 - 1
 *
 *
 */

// @lc code=start
/**
 * 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 isValid(root, nullptr, nullptr);
    }
    bool isValid(TreeNode* root, TreeNode* left, TreeNode* right) {
        if(!root) return true;

        if((left && left->val >= root->val) || right && root->val >= right->val)
            return false;

        return isValid(root->left, left, root) && isValid(root->right, root, right);
    }
};
// @lc code=end

• T: $O(N)$
• S: $O(N)$