Twin Trees - Amazon Top Interview Questions


Problem Statement :


Given two binary trees, root0 and root1, return whether their structure and values are equal.

Constraints

n ≤ 100,000 where n is the number of nodes in root0

m ≤ 100,000 where m is the number of nodes in root1

Example 1

Input

root0 = [0, [5, null, null], [9, null, null]]

root1 = [0, [5, null, null], [9, null, null]]

Output

True

Explanation

These two trees have the same values and same structure.

Example 2

Input

root0 = [0, [5, null, null], [9, null, null]]

root1 = [1, [2, null, null], [3, null, null]]

Output

False

Explanation

These two trees are not twins since their values are different.

Example 3

Input

root0 = [0, [5, null, null], null]

root1 = [0, null, [5, null, null]]

Output

False

Explanation

These two trees are not twins since their structure is different.


Solution :



title-img 




                        Solution in C++ :

bool solve(Tree* root0, Tree* root1) {
    if (root0 == NULL and root1 == NULL)
        return true;
    else if (root0 == NULL ^ root1 == NULL)
        return false;
    return (root0->val == root1->val and solve(root0->left, root1->left) and
            solve(root0->right, root1->right));
}
                    


                        Solution in Java :

import java.util.*;
class Solution {
    public boolean solve(Tree root0, Tree root1) {
        if (root0 == null && root1 == null) {
            return true;
        } else if (root0 == null ^ root1 == null) {
            return false;
        }
        return root0.val == root1.val && solve(root0.left, root1.left)
            && solve(root0.right, root1.right);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, root0, root1):
        if not root0 or not root1:
            return not root0 and not root1

        if root0.val != root1.val:
            return False

        return self.solve(root0.left, root1.left) and self.solve(root0.right, root1.right)


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