Check if given Preorder, Inorder and Postorder traversals are of same tree | Set 2
Given Preorder, Inorder and Postorder traversals of some tree. The task is 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 : NoWe already have discussed an approach to solve the above problem by constructing a tree using any two traversals in the previous article.
In this article, an approach without using any extra space is discussed.
Approach:
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- Search for the first element of preorder array in the inorder array and store it’s index as idx, if it doesn’t exist then return False. Also, check if that element present in the postorder array or not. If it is not then return False.
- Everything from 0th index for inorder and postorder and from 1st index for preorder of length idx becomes left subtree for first element of the preorder array.
- Everything from position idx+1 for inorder and preorder and from idx for postorder of length (length-idx-1) becomes right subtree for first element of preorder array.
- Repeat the steps 1 to 3 recursively until length of arrays become either 0 (in which case we
return true) or 1 (in which case we return True only if all three arrays are equal, else False).
Below is the implementation of the above approach:
C++
// C++ program to check if the given// three traversals are of the same// tree or not#include <bits/stdc++.h>using namespace std;// Function to check if traversals are// of the same treeint checktree(int preorder[], int inorder[], int postorder[], int len){ // if the array lengths are 0, // then all of them are obviously equal if (len == 0) return 1; // if array lengths are 1, // then check if all of them are equal if (len == 1) return (preorder[0] == inorder[0]) && (inorder[0] == postorder[0]); // search for first element of preorder // in inorder array int idx = -1, f = 0; for (int i = 0; i < len; ++i) if (inorder[i] == preorder[0]) { idx = i; break; } if(idx != -1){ for(int i = 0; i < len; i++) if(preorder[0] == postorder[i]){f = 1; break;} } if (idx == -1 || f == 0) return 0; // check for the left subtree int ret1 = checktree(preorder + 1, inorder, postorder, idx); // check for the right subtree int ret2 = checktree(preorder + idx + 1, inorder + idx + 1, postorder + idx, len - idx - 1); // return 1 only if both of them are // correct else 0 return (ret1 && ret2);}// Driver Codeint main(){ // Traversal Arrays int inorder[] = { 4, 2, 5, 1, 3 }; int preorder[] = { 1, 2, 4, 5, 3 }; int postorder[] = { 4, 5, 2, 3, 1 }; int len1 = sizeof(inorder) / sizeof(inorder[0]); int len2 = sizeof(preorder) / sizeof(preorder[0]); int len3 = sizeof(postorder) / sizeof(postorder[0]); // Check if all the array lengths are equal if ((len1 == len2) && (len2 == len3)) { bool res = checktree(preorder, inorder, postorder, len1); (res) ? cout << "Yes" : cout << "No"; } else cout << "No\n"; return 0;} |
Java
// Java program to check if the given// three traversals are of the same// tree or notclass GFG { // Function to check if traversals are // of the same tree static boolean checktree(int preorder[], int s, int inorder[], int s1, int postorder[], int s2, int len) { // if the array lengths are 0, // then all of them are obviously equal if (len == 0) return true; // if array lengths are 1, // then check if all of them are equal if (len == 1) return ((preorder[s] == inorder[s1]) && (inorder[s1] == postorder[s2])); // search for first element of preorder // in inorder array int idx = -1; for (int i = s1; i < s1 + len; ++i) if (inorder[i] == preorder[s]) { idx = i; break; } if (idx == -1) return false; idx = idx - s1; // check for the left subtree boolean ret1 = checktree(preorder, s + 1, inorder, s1, postorder, s2, idx); // check for the right subtree boolean ret2 = checktree( preorder, s + idx + 1, inorder, s1 + idx + 1, postorder, s2 + idx, len - idx - 1); // return 1 only if both of them are // correct else 0 return (ret1 && ret2); } // Driver Code public static void main(String args[]) { // Traversal Arrays int inorder[] = { 4, 2, 5, 1, 3 }; int preorder[] = { 1, 2, 4, 5, 3 }; int postorder[] = { 4, 5, 2, 3, 1 }; int len1 = inorder.length; int len2 = preorder.length; int len3 = postorder.length; // Check if all the array lengths are equal if ((len1 == len2) && (len2 == len3)) { boolean res = checktree(preorder, 0, inorder, 0, postorder, 0, len1); System.out.print(((res) ? "Yes" : "No")); } else System.out.print("No\n"); }}// This code is contributed by Arnab Kundu |
Python
# Python program to check if the given# three traversals are of the same# tree or not# Function to check if all three traversals# are of the same treedef checktree(preorder, inorder, postorder, length): # if the array lengths are 0, # then all of them are obviously equal if length == 0: return 1 # if array lengths are 1, # then check if all of them are equal if length == 1: return (preorder[0] == inorder[0]) and (inorder[0] == postorder[0]) # search for first element of preorder # in inorder array idx = -1 for i in range(length): if inorder[i] == preorder[0]: idx = i break if idx == -1: return 0 # check for the left subtree ret1 = checktree(preorder[1:], inorder, postorder, idx) # check for the right subtree ret2 = checktree(preorder[idx + 1:], inorder[idx + 1:], postorder[idx:], length-idx-1) # return 1 only if both of them are correct else 0 return (ret1 and ret2)# Driver Codeif __name__ == "__main__": inorder = [4, 2, 5, 1, 3] preorder = [1, 2, 4, 5, 3] postorder = [4, 5, 2, 3, 1] len1 = len(inorder) len2 = len(preorder) len3 = len(postorder) # check if all the array lengths are equal if (len1 == len2) and (len2 == len3): correct = checktree(preorder, inorder, postorder, len1) if (correct): print("Yes") else: print("No") else: print("No") |
C#
// C# program to check if the given// three traversals are of the same// tree or notusing System;class GFG { // Function to check if traversals are // of the same tree static bool checktree(int[] preorder, int s, int[] inorder, int s1, int[] postorder, int s2, int len) { // if the array lengths are 0, // then all of them are obviously equal if (len == 0) return true; // if array lengths are 1, // then check if all of them are equal if (len == 1) return ((preorder[s] == inorder[s1]) && (inorder[s1] == postorder[s2])); // search for first element of preorder // in inorder array int idx = -1; for (int i = s1; i < len; ++i) if (inorder[i] == preorder[s]) { idx = i; break; } if (idx == -1) return false; // check for the left subtree bool ret1 = checktree(preorder, s + 1, inorder, s1, postorder, s2, idx); // check for the right subtree bool ret2 = checktree( preorder, s + idx + 1, inorder, s1 + idx + 1, postorder, s2 + idx, len - idx - 1); // return 1 only if both of them are // correct else 0 return (ret1 && ret2); } // Driver Code static public void Main() { // Traversal Arrays int[] inorder = { 4, 2, 5, 1, 3 }; int[] preorder = { 1, 2, 4, 5, 3 }; int[] postorder = { 4, 5, 2, 3, 1 }; int len1 = inorder.Length; int len2 = preorder.Length; int len3 = postorder.Length; // Check if all the array lengths are equal if ((len1 == len2) && (len2 == len3)) { bool res = checktree(preorder, 0, inorder, 0, postorder, 0, len1); Console.Write(((res) ? "Yes" : "No")); } else Console.Write("No\n"); }}// This code is contributed by ajit |
PHP
<?php// PHP program to check if the given// three traversals are of the same// tree or not// Function to check if traversals are// of the same treefunction checktree($preorder, $inorder, $postorder, $len){ // if the array lengths are 0, // then all of them are obviously equal if ($len == 0) return 1; // if array lengths are 1, // then check if all of them are equal if ($len == 1) return ($preorder[0] == $inorder[0]) && ($inorder[0] == $postorder[0]); // search for first element of preorder // in inorder array $idx = -1; for ($i = 0; $i < $len; ++$i) if ($inorder[$i] == $preorder[0]) { $idx = $i; break; } if ($idx == -1) return 0; // check for the left subtree $ret1 = checktree(array_slice($preorder,1), $inorder, $postorder, $idx); // check for the right subtree $ret2 = checktree(array_slice($preorder,$idx + 1), array_slice($inorder,$idx + 1), array_slice($postorder,$idx), $len - $idx - 1); // return 1 only if both of them are // correct else 0 return ($ret1 && $ret2);} // Driver Code // Traversal Arrays $inorder = array( 4, 2, 5, 1, 3 ); $preorder = array( 1, 2, 4, 5, 3 ); $postorder = array( 4, 5, 2, 3, 1 ); $len1 = count($inorder) ; $len2 = count($preorder) ; $len3 = count($postorder) ; // Check if all the array lengths are equal if (($len1 == $len2) && ($len2 == $len3)) { $res = checktree($preorder, $inorder, $postorder, $len1); if($res) echo "Yes"; else echo "No"; } else echo "No\n"; // This code is contributed by AnkitRai01?> |
Javascript
<script> // Javascript program to check if the given // three traversals are of the same // tree or not // Function to check if traversals are // of the same tree function checktree(preorder, s, inorder, s1, postorder, s2, len) { // if the array lengths are 0, // then all of them are obviously equal if (len == 0) return true; // if array lengths are 1, // then check if all of them are equal if (len == 1) return ((preorder[s] == inorder[s1]) && (inorder[s1] == postorder[s2])); // search for first element of preorder // in inorder array let idx = -1; for (let i = s1; i < len; ++i) if (inorder[i] == preorder[s]) { idx = i; break; } if (idx == -1) return false; // check for the left subtree let ret1 = checktree(preorder, s + 1, inorder, s1, postorder, s2, idx); // check for the right subtree let ret2 = checktree( preorder, s + idx + 1, inorder, s1 + idx + 1, postorder, s2 + idx, len - idx - 1); // return 1 only if both of them are // correct else 0 return (ret1 && ret2); } // Traversal Arrays let inorder = [ 4, 2, 5, 1, 3 ]; let preorder = [ 1, 2, 4, 5, 3 ]; let postorder = [ 4, 5, 2, 3, 1 ]; let len1 = inorder.length; let len2 = preorder.length; let len3 = postorder.length; // Check if all the array lengths are equal if ((len1 == len2) && (len2 == len3)) { let res = checktree(preorder, 0, inorder, 0, postorder, 0, len1); document.write(((res) ? "Yes" : "No")); } else document.write("No");</script> |
Output:
Yes
