Check if given Preorder, Inorder and Postorder traversals are of same tree | Set 2

Given Preorder, Inorder and Postorder traversals of some tree. The task is to check if they all are of the same tree.
Examples:

Input : Inorder -> 4 2 5 1 3
        Preorder -> 1 2 4 5 3
        Postorder -> 4 5 2 3 1
Output : Yes
Explanation : All of the above three traversals are of 
the same tree.             1
                         /   \
                        2     3
                      /   \
                     4     5

Input : Inorder -> 4 2 5 1 3
        Preorder -> 1 5 4 2 3
        Postorder -> 4 1 2 3 5
Output : No

We already have discussed an approach to solve the above problem by constructing a tree using any two traversals in the previous article. 
In this article, an approach without using any extra space is discussed.
Approach: 

1. Search for the first element of preorder array in the inorder array and store it’s index as idx, if it doesn’t exist then return False. Also, check if that element present in the postorder array or not. If it is not then return False.
2. Everything from 0th index for inorder and postorder and from 1st index for preorder of length idx becomes left subtree for first element of the preorder array.
3. Everything from position idx+1 for inorder and preorder and from idx for postorder of length (length-idx-1) becomes right subtree for first element of preorder array.
4. Repeat the steps 1 to 3 recursively until length of arrays become either 0 (in which case we 
   return true) or 1 (in which case we return True only if all three arrays are equal, else False).

Below is the implementation of the above approach: 

C++

// C++ program to check if the given
// three traversals are of the same
// tree or not

#include <bits/stdc++.h>
using namespace std;

// Function to check if traversals are
// of the same tree
int checktree(int preorder[], int inorder[],
              int postorder[], int len)
{

    // if the array lengths are 0,
    // then all of them are obviously equal
    if (len == 0)
        return 1;

    // if array lengths are 1,
    // then check if all of them are equal
    if (len == 1)
        return (preorder[0] == inorder[0])
               && (inorder[0] == postorder[0]);

    // search for first element of preorder
    // in inorder array
    int idx = -1, f = 0;
    for (int i = 0; i < len; ++i)
        if (inorder[i] == preorder[0]) {
            idx = i;
            break;
        }

    if(idx != -1){
      for(int i = 0; i < len; i++)
        if(preorder[0] == postorder[i]){f = 1; break;}
    }

    if (idx == -1 || f == 0)
        return 0;

    // check for the left subtree
    int ret1
        = checktree(preorder + 1, inorder, postorder, idx);

    // check for the right subtree
    int ret2
        = checktree(preorder + idx + 1, inorder + idx + 1,
                    postorder + idx, len - idx - 1);

    // return 1 only if both of them are
    // correct else 0
    return (ret1 && ret2);
}

// Driver Code
int main()
{
    // Traversal Arrays
    int inorder[] = { 4, 2, 5, 1, 3 };
    int preorder[] = { 1, 2, 4, 5, 3 };
    int postorder[] = { 4, 5, 2, 3, 1 };
    int len1 = sizeof(inorder) / sizeof(inorder[0]);
    int len2 = sizeof(preorder) / sizeof(preorder[0]);
    int len3 = sizeof(postorder) / sizeof(postorder[0]);

    // Check if all the array lengths are equal
    if ((len1 == len2) && (len2 == len3)) {

        bool res
            = checktree(preorder, inorder, postorder, len1);

        (res) ? cout << "Yes" : cout << "No";
    }
    else
        cout << "No\n";

    return 0;
}

Java

// Java program to check if the given
// three traversals are of the same
// tree or not
class GFG {

    // Function to check if traversals are
    // of the same tree
    static boolean checktree(int preorder[], int s,
                             int inorder[], int s1,
                             int postorder[], int s2,
                             int len)
    {

        // if the array lengths are 0,
        // then all of them are obviously equal
        if (len == 0)
            return true;

        // if array lengths are 1,
        // then check if all of them are equal
        if (len == 1)
            return ((preorder[s] == inorder[s1])
                    && (inorder[s1] == postorder[s2]));

        // search for first element of preorder
        // in inorder array
        int idx = -1;
        for (int i = s1; i < s1 + len; ++i)
            if (inorder[i] == preorder[s]) {
                idx = i;
                break;
            }

        if (idx == -1)
            return false;
        idx = idx - s1;

        // check for the left subtree
        boolean ret1 = checktree(preorder, s + 1, inorder,
                                 s1, postorder, s2, idx);

        // check for the right subtree
        boolean ret2 = checktree(
            preorder, s + idx + 1, inorder, s1 + idx + 1,
            postorder, s2 + idx, len - idx - 1);

        // return 1 only if both of them are
        // correct else 0
        return (ret1 && ret2);
    }

    // Driver Code
    public static void main(String args[])
    {
        // Traversal Arrays
        int inorder[] = { 4, 2, 5, 1, 3 };
        int preorder[] = { 1, 2, 4, 5, 3 };
        int postorder[] = { 4, 5, 2, 3, 1 };
        int len1 = inorder.length;
        int len2 = preorder.length;
        int len3 = postorder.length;

        // Check if all the array lengths are equal
        if ((len1 == len2) && (len2 == len3)) {

            boolean res = checktree(preorder, 0, inorder, 0,
                                    postorder, 0, len1);

            System.out.print(((res) ? "Yes" : "No"));
        }
        else
            System.out.print("No\n");
    }
}

// This code is contributed by Arnab Kundu

Python

# Python program to check if the given
# three traversals are of the same
# tree or not

# Function to check if all three traversals
# are of the same tree


def checktree(preorder, inorder, postorder, length):

    # if the array lengths are 0,
    # then all of them are obviously equal
    if length == 0:
        return 1

    # if array lengths are 1,
    # then check if all of them are equal
    if length == 1:
        return (preorder[0] == inorder[0]) and (inorder[0] == postorder[0])

    # search for first element of preorder
    # in inorder array
    idx = -1

    for i in range(length):
        if inorder[i] == preorder[0]:
            idx = i
            break

    if idx == -1:
        return 0

    # check for the left subtree
    ret1 = checktree(preorder[1:], inorder, postorder, idx)

    # check for the right subtree
    ret2 = checktree(preorder[idx + 1:], inorder[idx + 1:],
                     postorder[idx:], length-idx-1)

    # return 1 only if both of them are correct else 0
    return (ret1 and ret2)


# Driver Code
if __name__ == "__main__":
    inorder = [4, 2, 5, 1, 3]
    preorder = [1, 2, 4, 5, 3]
    postorder = [4, 5, 2, 3, 1]
    len1 = len(inorder)
    len2 = len(preorder)
    len3 = len(postorder)

    # check if all the array lengths are equal
    if (len1 == len2) and (len2 == len3):
        correct = checktree(preorder, inorder,
                            postorder, len1)
        if (correct):
            print("Yes")
        else:
            print("No")
    else:
        print("No")

C#

// C# program to check if the given
// three traversals are of the same
// tree or not
using System;

class GFG {

    // Function to check if traversals are
    // of the same tree
    static bool checktree(int[] preorder, int s,
                          int[] inorder, int s1,
                          int[] postorder, int s2, int len)
    {

        // if the array lengths are 0,
        // then all of them are obviously equal
        if (len == 0)
            return true;

        // if array lengths are 1,
        // then check if all of them are equal
        if (len == 1)
            return ((preorder[s] == inorder[s1])
                    && (inorder[s1] == postorder[s2]));

        // search for first element of preorder
        // in inorder array
        int idx = -1;
        for (int i = s1; i < len; ++i)
            if (inorder[i] == preorder[s]) {
                idx = i;
                break;
            }

        if (idx == -1)
            return false;

        // check for the left subtree
        bool ret1 = checktree(preorder, s + 1, inorder, s1,
                              postorder, s2, idx);

        // check for the right subtree
        bool ret2 = checktree(
            preorder, s + idx + 1, inorder, s1 + idx + 1,
            postorder, s2 + idx, len - idx - 1);

        // return 1 only if both of them are
        // correct else 0
        return (ret1 && ret2);
    }

    // Driver Code
    static public void Main()
    {
        // Traversal Arrays
        int[] inorder = { 4, 2, 5, 1, 3 };
        int[] preorder = { 1, 2, 4, 5, 3 };
        int[] postorder = { 4, 5, 2, 3, 1 };
        int len1 = inorder.Length;
        int len2 = preorder.Length;
        int len3 = postorder.Length;

        // Check if all the array lengths are equal
        if ((len1 == len2) && (len2 == len3)) {

            bool res = checktree(preorder, 0, inorder, 0,
                                 postorder, 0, len1);

            Console.Write(((res) ? "Yes" : "No"));
        }
        else
            Console.Write("No\n");
    }
}

// This code is contributed by ajit

PHP

<?php
// PHP program to check if the given
// three traversals are of the same
// tree or not


// Function to check if traversals are
// of the same tree
function checktree($preorder, $inorder,
                $postorder, $len)
{

    // if the array lengths are 0,
    // then all of them are obviously equal
    if ($len == 0)
        return 1;

    // if array lengths are 1,
    // then check if all of them are equal
    if ($len == 1)
        return ($preorder[0] == $inorder[0])
            && ($inorder[0] == $postorder[0]);

    // search for first element of preorder
    // in inorder array
    $idx = -1;
    for ($i = 0; $i < $len; ++$i)
        if ($inorder[$i] == $preorder[0])
        {
            $idx = $i;
            break;
        }

    if ($idx == -1)
        return 0;

    // check for the left subtree
    $ret1 = checktree(array_slice($preorder,1), $inorder,
                        $postorder, $idx);

    // check for the right subtree
    $ret2 = checktree(array_slice($preorder,$idx + 1),
                        array_slice($inorder,$idx + 1),
                        array_slice($postorder,$idx), $len - $idx - 1);

    // return 1 only if both of them are
    // correct else 0
    return ($ret1 && $ret2);
}

    // Driver Code

    // Traversal Arrays
    $inorder = array( 4, 2, 5, 1, 3 );
    $preorder = array( 1, 2, 4, 5, 3 );
    $postorder = array( 4, 5, 2, 3, 1 );
    $len1 = count($inorder) ;
    $len2 = count($preorder) ;
    $len3 = count($postorder) ;

    // Check if all the array lengths are equal
    if (($len1 == $len2) && ($len2 == $len3))
    {

        $res = checktree($preorder, $inorder,
                            $postorder, $len1);

        if($res)
            echo "Yes";
        else
            echo "No";
    }
    else
        echo "No\n";

    // This code is contributed by AnkitRai01
?>

Javascript

<script>
    // Javascript program to check if the given
    // three traversals are of the same
    // tree or not

    // Function to check if traversals are
    // of the same tree
    function checktree(preorder, s, inorder, s1, postorder, s2, len)
    {

        // if the array lengths are 0,
        // then all of them are obviously equal
        if (len == 0)
            return true;

        // if array lengths are 1,
        // then check if all of them are equal
        if (len == 1)
            return ((preorder[s] == inorder[s1])
                    && (inorder[s1] == postorder[s2]));

        // search for first element of preorder
        // in inorder array
        let idx = -1;
        for (let i = s1; i < len; ++i)
            if (inorder[i] == preorder[s]) {
                idx = i;
                break;
            }

        if (idx == -1)
            return false;

        // check for the left subtree
        let ret1 = checktree(preorder, s + 1, inorder, s1,
                              postorder, s2, idx);

        // check for the right subtree
        let ret2 = checktree(
            preorder, s + idx + 1, inorder, s1 + idx + 1,
            postorder, s2 + idx, len - idx - 1);

        // return 1 only if both of them are
        // correct else 0
        return (ret1 && ret2);
    }

    // Traversal Arrays
    let inorder = [ 4, 2, 5, 1, 3 ];
    let preorder = [ 1, 2, 4, 5, 3 ];
    let postorder = [ 4, 5, 2, 3, 1 ];
    let len1 = inorder.length;
    let len2 = preorder.length;
    let len3 = postorder.length;

    // Check if all the array lengths are equal
    if ((len1 == len2) && (len2 == len3)) {

      let res = checktree(preorder, 0, inorder, 0,
                           postorder, 0, len1);

      document.write(((res) ? "Yes" : "No"));
    }
    else
      document.write("No");

</script>

Output:

Yes

Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N), as we are using a recursive stack for the checktree function.