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

Given Preorder , Inorder and Postorder traversals of some tree. Write a program 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
Exaplanation : 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The most basic approach to solve this problem will be to first construct a tree using two of the three given traversals and then do the third traversal on this constructed tree and compare it with the given traversal. If both of the traversals are same then print Yes otherwise print No. Here, we use Inorder and Preorder traversals to construct the tree. We may also use Inorder and Postorder traversal instead of Preorder traversal for tree construction. You may refer to this post on how to construct tree from given Inorder and Preorder traversal. After constructing the tree, we will obtain the Postorder traversal of this tree and compare it with the given Postorder traversal.

Below is the implementation of above approach:

C++

/* C++ program to check if all three given 
   traversals are of the same tree */
#include <bits/stdc++.h> 
using namespace std; 
  
// A Binary Tree Node 
struct Node 
{ 
    int data; 
    struct Node *left, *right; 
}; 
  
// Utility function to create a new tree node 
Node* newNode(int data) 
{ 
    Node *temp = new Node; 
    temp->data = data; 
    temp->left = temp->right = NULL; 
    return temp; 
} 
  
/* Function to find index of value in arr[start...end] 
   The function assumes that value is present in in[] */
int search(int arr[], int strt, int end, int value) 
{ 
    for (int i = strt; i <= end; i++) 
    { 
        if(arr[i] == value) 
            return i; 
    } 
} 
  
/* Recursive function to construct binary tree  
   of size len from Inorder traversal in[] and  
   Preorder traversal pre[].  Initial values 
   of inStrt and inEnd should be 0 and len -1.   
   The function doesn't do any error checking for  
   cases where inorder and preorder do not form a  
   tree */
Node* buildTree(int in[], int pre[], int inStrt,  
                                      int inEnd) 
{ 
    static int preIndex = 0; 
   
    if(inStrt > inEnd) 
        return NULL; 
   
    /* Pick current node from Preorder traversal  
       using preIndex and increment preIndex */
    Node *tNode = newNode(pre[preIndex++]); 
   
    /* If this node has no children then return */
    if (inStrt == inEnd) 
        return tNode; 
   
    /* Else find the index of this node in  
       Inorder traversal */
    int inIndex = search(in, inStrt, inEnd, tNode->data); 
   
    /* Using index in Inorder traversal,  
       construct left and right subtress */
    tNode->left = buildTree(in, pre, inStrt, inIndex-1); 
    tNode->right = buildTree(in, pre, inIndex+1, inEnd); 
   
    return tNode; 
} 
  
/* function to compare Postorder traversal  
   on constructed tree and given Postorder */
int checkPostorder(Node* node, int postOrder[], int index) 
{ 
    if (node == NULL) 
        return index; 
   
    /* first recur on left child */
    index = checkPostorder(node->left,postOrder,index); 
       
    /* now recur on right child */
    index = checkPostorder(node->right,postOrder,index);     
    
    /* Compare if data at current index in  
       both Postorder traversals are same */
    if (node->data == postOrder[index]) 
        index++; 
    else
        return -1; 
  
    return index; 
} 
  
// Driver program to test above functions 
int main() 
{ 
    int inOrder[] = {4, 2, 5, 1, 3}; 
    int preOrder[] = {1, 2, 4, 5, 3}; 
    int postOrder[] = {4, 5, 2, 3, 1}; 
  
    int len = sizeof(inOrder)/sizeof(inOrder[0]); 
  
    // build tree from given  
    // Inorder and Preorder traversals 
    Node *root = buildTree(inOrder, preOrder, 0, len - 1); 
  
    // compare postorder traversal on constructed 
    // tree with given Postorder traversal 
    int index = checkPostorder(root,postOrder,0); 
  
    // If both postorder traversals are same  
    if (index == len) 
        cout << "Yes"; 
    else
        cout << "No"; 
  
    return 0; 
} 

Java

/* Java program to check if all three given  
traversals are of the same tree */
import java.util.*; 
class GfG { 
    static int preIndex = 0; 
  
// A Binary Tree Node  
static class Node  
{  
    int data;  
    Node left, right;  
} 
  
// Utility function to create a new tree node  
static Node newNode(int data)  
{  
    Node temp = new Node();  
    temp.data = data;  
    temp.left = null; 
    temp.right = null;  
    return temp;  
}  
  
/* Function to find index of value in arr[start...end]  
The function assumes that value is present in in[] */
static int search(int arr[], int strt, int end, int value)  
{  
    for (int i = strt; i <= end; i++)  
    {  
        if(arr[i] == value)  
            return i;  
    }  
    return -1; 
}  
  
/* Recursive function to construct binary tree  
of size len from Inorder traversal in[] and  
Preorder traversal pre[]. Initial values  
of inStrt and inEnd should be 0 and len -1.  
The function doesn't do any error checking for  
cases where inorder and preorder do not form a  
tree */
static Node buildTree(int in[], int pre[], int inStrt, int inEnd)  
{  
  
    if(inStrt > inEnd)  
        return null;  
  
    /* Pick current node from Preorder traversal  
    using preIndex and increment preIndex */
    Node tNode = newNode(pre[preIndex++]);  
  
    /* If this node has no children then return */
    if (inStrt == inEnd)  
        return tNode;  
  
    /* Else find the index of this node in  
    Inorder traversal */
    int inIndex = search(in, inStrt, inEnd, tNode.data);  
  
    /* Using index in Inorder traversal,  
    construct left and right subtress */
    tNode.left = buildTree(in, pre, inStrt, inIndex-1);  
    tNode.right = buildTree(in, pre, inIndex+1, inEnd);  
  
    return tNode;  
}  
  
/* function to compare Postorder traversal  
on constructed tree and given Postorder */
static int checkPostorder(Node node, int postOrder[], int index)  
{  
    if (node == null)  
        return index;  
  
    /* first recur on left child */
    index = checkPostorder(node.left,postOrder,index);  
      
    /* now recur on right child */
    index = checkPostorder(node.right,postOrder,index);      
      
    /* Compare if data at current index in  
    both Postorder traversals are same */
    if (node.data == postOrder[index])  
        index++;  
    else
        return -1;  
  
    return index;  
}  
  
// Driver program to test above functions  
public static void main(String[] args)  
{  
    int inOrder[] = {4, 2, 5, 1, 3};  
    int preOrder[] = {1, 2, 4, 5, 3};  
    int postOrder[] = {4, 5, 2, 3, 1};  
  
    int len = inOrder.length;  
  
    // build tree from given  
    // Inorder and Preorder traversals  
    Node root = buildTree(inOrder, preOrder, 0, len - 1);  
  
    // compare postorder traversal on constructed  
    // tree with given Postorder traversal  
    int index = checkPostorder(root,postOrder,0);  
  
    // If both postorder traversals are same  
    if (index == len)  
        System.out.println("Yes");  
    else
        System.out.println("No");  
  
} 
}  

C#

/* C# program to check if all three given  
traversals are of the same tree */
using System; 
      
public class GfG  
{ 
    static int preIndex = 0; 
  
    // A Binary Tree Node  
    class Node  
    {  
        public int data;  
        public Node left, right;  
    } 
  
    // Utility function to create a new tree node  
    static Node newNode(int data)  
    {  
        Node temp = new Node();  
        temp.data = data;  
        temp.left = null; 
        temp.right = null;  
        return temp;  
    }  
  
    /* Function to find index of  
    value in arr[start...end]  
    The function assumes that  
    value is present in in[] */
    static int search(int []arr, int strt, 
                        int end, int value)  
    {  
        for (int i = strt; i <= end; i++)  
        {  
            if(arr[i] == value)  
                return i;  
        }  
        return -1; 
    }  
  
    /* Recursive function to construct  
    binary tree of size len from Inorder 
    traversal in[] and Preorder traversal 
    pre[]. Initial values of inStrt and  
    inEnd should be 0 and len -1. The  
    function doesn't do any error checking for  
    cases where inorder and preorder do not form a  
    tree */
    static Node buildTree(int []In, int []pre, 
                            int inStrt, int inEnd)  
    {  
  
        if(inStrt > inEnd)  
            return null;  
  
        /* Pick current node from Preorder traversal  
        using preIndex and increment preIndex */
        Node tNode = newNode(pre[preIndex++]);  
  
        /* If this node has no children then return */
        if (inStrt == inEnd)  
            return tNode;  
  
        /* Else find the index of this node in  
        Inorder traversal */
        int inIndex = search(In, inStrt, inEnd, tNode.data);  
  
        /* Using index in Inorder traversal,  
        construct left and right subtress */
        tNode.left = buildTree(In, pre, inStrt, inIndex - 1);  
        tNode.right = buildTree(In, pre, inIndex + 1, inEnd);  
  
        return tNode;  
    }  
  
    /* function to compare Postorder traversal  
    on constructed tree and given Postorder */
    static int checkPostorder(Node node, int []postOrder, int index)  
    {  
        if (node == null)  
            return index;  
  
        /* first recur on left child */
        index = checkPostorder(node.left,postOrder,index);  
  
        /* now recur on right child */
        index = checkPostorder(node.right,postOrder,index);   
  
        /* Compare if data at current index in  
        both Postorder traversals are same */
        if (node.data == postOrder[index])  
            index++;  
        else
            return -1;  
  
        return index;  
    }  
  
    // Driver code  
    public static void Main()  
    {  
        int []inOrder = {4, 2, 5, 1, 3};  
        int []preOrder = {1, 2, 4, 5, 3};  
        int []postOrder = {4, 5, 2, 3, 1};  
  
        int len = inOrder.Length;  
  
        // build tree from given  
        // Inorder and Preorder traversals  
        Node root = buildTree(inOrder, preOrder, 0, len - 1);  
  
        // compare postorder traversal on constructed  
        // tree with given Postorder traversal  
        int index = checkPostorder(root, postOrder, 0);  
  
        // If both postorder traversals are same  
        if (index == len)  
            Console.WriteLine("Yes");  
        else
            Console.WriteLine("No");  
  
    } 
} 
  
/* This code is contributed PrinciRaj1992 */

Output:

Yes

Time Complexity : O( n * n ), where n is number of nodes in the tree.

This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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