# 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 : No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We 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:

1. 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.
2. 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.
3. 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.
4. 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++ program to check if the given // three traversals are of the same // tree or not    #include using namespace std;    // Function to check if traversals are // of the same tree int 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;     for (int i = 0; i < len; ++i)         if (inorder[i] == preorder[0]) {             idx = i;             break;         }        if (idx == -1)         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 Code int 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 program to check if the given // three traversals are of the same // tree or not class 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 < len; ++i)         if (inorder[i] == preorder[s])          {             idx = i;             break;         }        if (idx == -1)         return false;        // 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 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 tree def 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 Code if __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# program to check if the given // three traversals are of the same // tree or not using 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



Output:
Yes

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