In this article, first count of possible BST (Binary Search Trees)s is discussed, then the 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. We have discussed different approaches to find n’th Catalan number here.
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.
- Initialize list of BSTs as empty.
- For every number i where i varies from 1 to N, do following
- Create a new node with key as ‘i’, let this node be ‘node’
- Recursively construct list of all left subtrees.
- Recursively construct list of all right subtrees.
- Iterate for all left subtrees
- For current leftsubtree, iterate for all right subtrees
- Add current left and right subtrees to ‘node’ and add
- node’ to list.
Below is the implementation of the above idea.
C++
// A C++ program to construct all unique BSTs for keys from 1 to n#include <bits/stdc++.h>using namespace std;// node structurestruct node{ int key; struct node *left, *right;};// A utility function to create a new BST nodestruct 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 BSTvoid preorder(struct node *root){ if (root != NULL) { cout << root->key << " "; preorder(root->left); preorder(root->right); }}// function for constructing treesvector<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 functionsint 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;} |
Java
// A Java program to construct all unique BSTs for keys from 1 to nimport java.util.ArrayList;public class Main { // function for constructing trees static ArrayList<Node> constructTrees(int start, int end) { ArrayList<Node> list=new ArrayList<>(); /* if start > end then subtree will be empty so returning NULL in the list */ if (start > end) { list.add(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 */ ArrayList<Node> leftSubtree = constructTrees(start, i - 1); /* constructing right subtree */ ArrayList<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++) { Node left = leftSubtree.get(j); for (int k = 0; k < rightSubtree.size(); k++) { Node right = rightSubtree.get(k); Node node = new Node(i); // making value i as root node.left = left; // connect left subtree node.right = right; // connect right subtree list.add(node); // add this tree to list } } } return list; } // A utility function to do preorder traversal of BST static void preorder(Node root) { if (root != null) { System.out.print(root.key+" ") ; preorder(root.left); preorder(root.right); } } public static void main(String args[]) { ArrayList<Node> totalTreesFrom1toN = constructTrees(1, 3); /* Printing preorder traversal of all constructed BSTs */ System.out.println("Preorder traversals of all constructed BSTs are "); for (int i = 0; i < totalTreesFrom1toN.size(); i++) { preorder(totalTreesFrom1toN.get(i)); System.out.println(); } }}// node structure class Node { int key; Node left, right; Node(int data) { this.key=data; left=right=null; }}; //This code is contributed by Gaurav Tiwari |
Python3
# Python3 program to construct all unique# BSTs for keys from 1 to n # Binary Tree Node """ A utility function to create anew BST node """class newNode: # Construct to create a newNode def __init__(self, item): self.key=item self.left = None self.right = None# A utility function to do preorder # traversal of BST def preorder(root) : if (root != None) : print(root.key, end = " " ) preorder(root.left) preorder(root.right) # function for constructing trees def constructTrees(start, end): list = [] """ if start > end then subtree will be empty so returning None in the list """ if (start > end) : list.append(None) return list """ iterating through all values from start to end for constructing left and right subtree recursively """ for i in range(start, end + 1): """ constructing left subtree """ leftSubtree = constructTrees(start, i - 1) """ constructing right subtree """ rightSubtree = constructTrees(i + 1, end) """ now looping through all left and right subtrees and connecting them to ith root below """ for j in range(len(leftSubtree)) : left = leftSubtree[j] for k in range(len(rightSubtree)): right = rightSubtree[k] node = newNode(i) # making value i as root node.left = left # connect left subtree node.right = right # connect right subtree list.append(node) # add this tree to list return list# Driver Code if __name__ == '__main__': # Construct all possible BSTs totalTreesFrom1toN = constructTrees(1, 3) """ Printing preorder traversal of all constructed BSTs """ print("Preorder traversals of all", "constructed BSTs are") for i in range(len(totalTreesFrom1toN)): preorder(totalTreesFrom1toN[i]) print()# This code is contributed by# Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to construct all // unique BSTs for keys from 1 to n using System.Collections;using System;class GFG { // function for constructing trees static ArrayList constructTrees(int start, int end) { ArrayList list = new ArrayList(); /* if start > end then subtree will be empty so returning NULL in the list */ if (start > end) { list.Add(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 */ ArrayList leftSubtree = constructTrees(start, i - 1); /* constructing right subtree */ ArrayList 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.Count; j++) { Node left = (Node)leftSubtree[j]; for (int k = 0; k < rightSubtree.Count; k++) { Node right = (Node)rightSubtree[k]; // making value i as root Node node = new Node(i); // connect left subtree node.left = left; // connect right subtree node.right = right; // Add this tree to list list.Add(node); } } } return list; } // A utility function to do // preorder traversal of BST static void preorder(Node root) { if (root != null) { Console.Write(root.key+" ") ; preorder(root.left); preorder(root.right); } } // Driver code public static void Main(String []args) { ArrayList totalTreesFrom1toN = constructTrees(1, 3); /* Printing preorder traversal of all constructed BSTs */ Console.WriteLine("Preorder traversals of all" + "constructed BSTs are "); for (int i = 0; i < totalTreesFrom1toN.Count; i++) { preorder((Node)totalTreesFrom1toN[i]); Console.WriteLine(); } } // node structure public class Node { public int key; public Node left, right; public Node(int data) { this.key=data; left=right=null; } }; } // This code is contributed by Arnab Kundu |
Javascript
<script>// node structure class Node { constructor(data) { this.key = data; this.left = null; this.right = null; }}; // Javascript program to construct all // unique BSTs for keys from 1 to n // function for constructing trees function constructTrees(start, end) { var list = []; /* if start > end then subtree will be empty so returning NULL in the list */ if (start > end) { list.push(null); return list; } /* iterating through all values from start to end for constructing\ left and right subtree recursively */ for (var i = start; i <= end; i++) { /* constructing left subtree */ var leftSubtree = constructTrees(start, i - 1); /* constructing right subtree */ var rightSubtree = constructTrees(i + 1, end); /* now looping through all left and right subtrees and connecting them to ith root below */ for (var j = 0; j < leftSubtree.length; j++) { var left = leftSubtree[j]; for (var k = 0; k < rightSubtree.length; k++) { var right = rightSubtree[k]; // making value i as root var node = new Node(i); // connect left subtree node.left = left; // connect right subtree node.right = right; // push this tree to list list.push(node); } } } return list; } // A utility function to do // preorder traversal of BST function preorder(root) { if (root != null) { document.write(root.key+" ") ; preorder(root.left); preorder(root.right); } } // Driver codevar totalTreesFrom1toN = constructTrees(1, 3); /* Printing preorder traversal of all constructed BSTs */document.write("Preorder traversals of all" + "constructed BSTs are<br>"); for (var i = 0; i < totalTreesFrom1toN.length; i++) { preorder(totalTreesFrom1toN[i]); document.write("<br>");} </script> |
Output
Preorder traversals of all constructed BSTs are 1 2 3 1 3 2 2 1 3 3 1 2 3 2 1
Time Complexity: O(2^n).
The time complexity of this solution is exponential. This is because we are constructing all possible BSTs. Since at each level, we are generating two subproblems and at each level, there are O(2^n) subproblems, the time complexity is O(2^n).
Auxiliary Space: O(2^n).
The space complexity of this solution is also exponential. This is because we are constructing all possible BSTs. Since at each level, we are generating two subproblems and at each level, there are O(2^n) subproblems, the space complexity is O(2^n).
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
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