Construct all possible BSTs for keys 1 to N - GeeksforGeeks

Construct all possible BSTs for keys 1 to N - GeeksforGeeks
count of possible BST (Binary Search Trees)s is discussed, then construction of all possible BSTs.

How to construct all BST for keys 1..N?
The idea is to maintain a list of roots of all BSTs. Recursively construct all possible left and right subtrees. Create a tree for every pair of left and right subtree and add the tree to list. Below is detailed algorithm.

1) Initialize list of BSTs as empty.  
2) For every number i where i varies from 1 to N, do following
......a)  Create a new node with key as 'i', let this node be 'node'
......b)  Recursively construct list of all left subtrees.
......c)  Recursively construct list of all right subtrees.
3) Iterate for all left subtrees
   a) For current leftsubtree, iterate for all right subtrees
        Add current left and right subtrees to 'node' and add
        'node' to list.

vector<struct node *> constructTrees(int start, int end)

{

    vector<struct node *> list;

 

    /*  if start > end   then subtree will be empty so returning NULL

        in the list */

    if (start > end)

    {

        list.push_back(NULL);

        return list;

    }

 

    /*  iterating through all values from start to end  for constructing\

        left and right subtree recursively  */

    for (int i = start; i <= end; i++)

    {

        /*  constructing left subtree   */

        vector<struct node *> leftSubtree  = constructTrees(start, i - 1);

 

        /*  constructing right subtree  */

        vector<struct node *> rightSubtree = constructTrees(i + 1, end);

 

        /*  now looping through all left and right subtrees and connecting

            them to ith root  below  */

        for (int j = 0; j < leftSubtree.size(); j++)

        {

            struct node* left = leftSubtree[j];

            for (int k = 0; k < rightSubtree.size(); k++)

            {

                struct node * right = rightSubtree[k];

                struct node * node = newNode(i);// making value i as root

                node->left = left;              // connect left subtree

                node->right = right;            // connect right subtree

                list.push_back(node);           // add this tree to list

            }

        }

    }

    return list;

}

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