02. Inorder Traversal and BST

The problem can be found at the following link: Question Link

My Approach

Since it is a BST, its inorder traversal should always be a sorted array, hence the code just checks if the array is sorted or not

• Iterates through the array.

• Checks if each element is less than its immediate successor.

• Returns 1 if the entire array satisfies the condition, otherwise returns 0.

Time and Auxiliary Space Complexity

• Time Complexity: O(N), where N is the number elements in the array

• Auxiliary Space Complexity: O(1), as no extra space is used.

Code (C++)

class Solution{
    public:
    int isRepresentingBST(int arr[], int N)
    {
        for (int i=0;i<N-1;i++)
        {
            if (arr[i]<arr[i+1])
                continue;
            else return 0;
        }
        return 1;
    }
};

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