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98. Validate Binary Search Tree | Leetcode solution

 98. Validate Binary Search Tree


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, 104].
  • -231 <= Node.val <= 231 - 1


solution :

  i have solved it using the property of BST  that is inorder traversal of 
  binary search tree gives an array of sorted order.
/**
 * 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:
    void inorder(TreeNode* root , vector<int>&result)
    {   
        if(!root)
        {
            return;
        }
        inorder(root->left , result );
        result.push_back(root->val);
        inorder(root->right , result);

    }
    bool isValidBST(TreeNode* root) {
        vector<int>result;
        inorder(root , result);
        for(int i = 1 ; i  < result.size() ; i++)
        {
             if(result[i] <= result[i-1])
                 return false;
                 
        }
        return true;
        
    }
};


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