# Construct all possible BSTs for keys 1 to N

In this article, first count of possible BST (Binary Search Trees)s is discussed, then construction of all possible BSTs.

How many structurally unique BSTs for keys from 1..N?

```For example, for N = 2, there are 2 unique BSTs
1               2
\            /
2         1

For N = 3, there are 5 possible BSTs
1              3        3         2      1
\           /        /        /  \      \
3        2         1        1    3      2
/       /            \                    \
2      1               2                    3
```

We strongly recommend you to minimize your browser and try this yourself first.

We know that all node in left subtree are smaller than root and in right subtree are larger than root so if we have ith number as root, all numbers from 1 to i-1 will be in left subtree and i+1 to N will be in right subtree. If 1 to i-1 can form x different trees and i+1 to N can from y different trees then we will have x*y total trees when ith number is root and we also have N choices for root also so we can simply iterate from 1 to N for root and another loop for left and right subtree. If we take a closer look, we can notice that the count is basically n’th Catalan number.
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
'node' to list.
```

Below is C++ implementation of above idea.

`// A C++ prgroam to contrcut all unique BSTs for keys from 1 to n`
`#include <iostream>`
`#include<vector>`
`using` `namespace` `std;`
`//  node structure`
`struct` `node`
`{`
`    ``int` `key;`
`    ``struct` `node *left, *right;`
`};`
`// A utility function to create a new BST node`
`struct` `node *newNode(``int` `item)`
`{`
`    ``struct` `node *temp =  ``new` `node;`
`    ``temp->key = item;`
`    ``temp->left = temp->right = NULL;`
`    ``return` `temp;`
`}`
`// A utility function to do preorder traversal of BST`
`void` `preorder(``struct` `node *root)`
`{`
`    ``if` `(root != NULL)`
`    ``{`
`        ``cout << root->key << ``" "``;`
`        ``preorder(root->left);`
`        ``preorder(root->right);`
`    ``}`
`}`
`//  function for constructing trees`
`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;`
`}`
`// Driver Program to test above functions`
`int` `main()`
`{`
`    ``// Construct all possible BSTs`
`    ``vector<``struct` `node *> totalTreesFrom1toN = constructTrees(1, 3);`
`    ``/*  Printing preorder traversal of all constructed BSTs   */`
`    ``cout << ``"Preorder traversals of all constructed BSTs are \n"``;`
`    ``for` `(``int` `i = 0; i < totalTreesFrom1toN.size(); i++)`
`    ``{`
`        ``preorder(totalTreesFrom1toN[i]);`
`        ``cout << endl;`
`    ``}`
`    ``return` `0;`
`}`

Output:

```Preorder traversals of all constructed BSTs are
1 2 3
1 3 2
2 1 3
3 1 2
3 2 1

```

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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