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 

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++

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/* 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;
}

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Java

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/* 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"); 
  
}

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C#

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/* 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 */

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Output:

Yes

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

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