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++
/* 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.
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