Find all possible binary trees with given Inorder Traversal - GeeksforGeeks

Find all possible binary trees with given Inorder Traversal - GeeksforGeeks
Given an array that represents Inorder Traversal, find all possible Binary Trees with the given Inorder traversal and print their preorder traversals.

Let given inorder traversal be in[]. In the given traversal, all nodes in left subtree of in[i] must appear before it and in right subtree must appear after it. So when we consider in[i] as root, all elements from in[0] to in[i-1] will be in left subtree and in[i+1] to n-1 will be in right subtree. If in[0] to in[i-1] can form xdifferent trees and in[i+1] to in[n-1] can from y different trees then we will have x*y total trees when in[i] as root. So we simply iterate from 0 to n-1 for root. For every node in[i], recursively find different left and right subtrees. 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.

The idea is to maintain a list of roots of all Binary Trees. 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 Binary Trees as empty.  
2) For every element in[i] where i varies from 0 to n-1,
    do following
......a)  Create a new node with key as 'arr[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.

class Node {

    int data;

    Node left, right;

    public Node(int item) {

        data = item;

        left = null;

        right = null;

    }

}

/* Class to print Level Order Traversal */

class BinaryTree {

    Node root;

    // A utility function to do preorder traversal of BST

    void preOrder(Node node) {

        if (node != null) {

            System.out.print(node.data + " "    );

            preOrder(node.left);

            preOrder(node.right);

        }

    }

    // Function for constructing all possible trees with

    // given inorder traversal stored in an array from

    // arr[start] to arr[end]. This function returns a

    // vector of trees.

    Vector<Node> getTrees(int arr[], int start, int end) {

        // List to store all possible trees

        Vector<Node> trees= new Vector<Node>();

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

         returning NULL in the list */

        if (start > end) {

            trees.add(null);

            return trees;

        }

        /* 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<Node> ltrees = getTrees(arr, start, i - 1);

             

            /* Constructing right subtree */

            Vector<Node> rtrees = getTrees(arr, i + 1, end);

            /* Now looping through all left and right subtrees

             and connecting them to ith root below */

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

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

                    // Making arr[i] as root

                    Node node = new Node(arr[i]);

                    // Connecting left subtree

                    node.left = ltrees.get(j);

                    // Connecting right subtree

                    node.right = rtrees.get(k);

                    // Adding this tree to list

                    trees.add(node);

                }

            }

        }

        return trees;

    }

class Node:

    # Utility to create a new node

    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 is not None:

        print root.key,

        preorder(root.left)

        preorder(root.right)

# Function for constructing all possible trees with

# given inorder traversal stored in an array from

# arr[start] to arr[end]. This function returns a

# vector of trees.

def getTrees(arr , start , end):

    # List to store all possible trees

    trees = []

     

    """ if start > end then subtree will be empty so

    returning NULL in the list """

    if start > end :

        trees.append(None)

        return trees

     

    """ 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

        ltrees = getTrees(arr , start , i-1)

         

        # Constructing right subtree

        rtrees = getTrees(arr , i+1 , end)

         

        """ Looping through all left and right subtrees

        and connecting to ith root below"""

        for j in ltrees :

            for k in rtrees :

                # Making arr[i]  as root

                node  = Node(arr[i])

     

                # Connecting left subtree

                node.left = j 

                # Connecting right subtree

                node.right = k

                # Adding this tree to list

                trees.append(node)

    return trees

# Driver program to test above function

inp = [4 , 5, 7]

n = len(inp)

trees = getTrees(inp , 0 , n-1)

print "Preorder traversals of different possible\

 Binary Trees are "

for i in trees :

    preorder(i);

    print ""

Related: Construct all possible BSTs for keys 1 to N
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