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 a 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 the 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 Nodestruct Node{ int data; struct Node *left, *right;};// Utility function to create a new tree nodeNode* 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 functionsint 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"); }} |
Python3
# Python3 program to check if# all three given traversals # are of the same tree class node: def __init__(self, x): self.data = x self.left = None self.right = NonepreIndex = 0# Function to find index of value # in arr[start...end]. The function # assumes that value is present in indef search(arr, strt, end, value): for i in range(strt, end + 1): if(arr[i] == value): return i# Recursive function to construct# binary tree of size lenn from # Inorder traversal in and Preorder # traversal pre[]. Initial values # of inStrt and inEnd should be 0 # and lenn -1. The function doesn't # do any error checking for cases # where inorder and preorder do not # form a tree def buildTree(inn, pre, inStrt, inEnd): global preIndex if(inStrt > inEnd): return None # Pick current node from Preorder # traversal using preIndex and # increment preIndex tNode = node(pre[preIndex]) preIndex += 1 # If this node has no children # then return if (inStrt == inEnd): return tNode # Else find the index of this # node in Inorder traversal inIndex = search(inn, inStrt, inEnd, tNode.data) # Using index in Inorder traversal, # construct left and right subtress tNode.left = buildTree(inn, pre, inStrt, inIndex - 1) tNode.right = buildTree(inn, pre, inIndex + 1, inEnd) return tNode# function to compare Postorder traversal# on constructed tree and given Postorderdef checkPostorder(node, postOrder, index): if (node == None): 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 += 1 else: return - 1 return index# Driver codeif __name__ == '__main__': inOrder = [4, 2, 5, 1, 3] preOrder = [1, 2, 4, 5, 3] postOrder = [4, 5, 2, 3, 1] lenn = len(inOrder) # build tree from given # Inorder and Preorder traversals root = buildTree(inOrder, preOrder, 0, lenn - 1) # compare postorder traversal on # constructed tree with given # Postorder traversal index = checkPostorder(root, postOrder, 0) # If both postorder traversals are same if (index == lenn): print("Yes") else: print("No")# This code is contributed by Mohit Kumar 29 |
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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
