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.


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 C++ implementation of above approach:

/* 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])
        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";
        cout << "No";

    return 0;



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 or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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