class Solution {
public String mergeAlternately(String word1, String word2) {
StringBuilder merged = new StringBuilder();
int i = 0, j = 0;
while (i < word1.length() || j < word2.length()) {
if (i < word1.length()) merged.append(word1.charAt(i++));
if (j < word2.length()) merged.append(word2.charAt(j++));
}
return merged.toString();
}
}

class Solution {
public String gcdOfStrings(String str1, String str2) {
if(str1.length() < str2.length()){
return gcdOfStrings(str2,str1);
}else if(!str1.startsWith(str2)){
return "";
}else if(str2.length() == 0){
return str1;
}else{
return gcdOfStrings(str1.substring(str2.length()),str2);
}
}
}

For two strings s and t, we say "t divides s" if and only if s = t + t + t + ... + t + t (i.e., t is concatenated with itself one or more times).

Given two strings str1 and str2, return the largest string x such that x divides both str1 and str2.



Example 1:

Input: str1 = "ABCABC", str2 = "ABC"
Output: "ABC"
Example 2:

Input: str1 = "ABABAB", str2 = "ABAB"
Output: "AB"
Example 3:

Input: str1 = "LEET", str2 = "CODE"
Output: ""

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951. Flip Equivalent Binary Trees
Solved
Medium
Topics
Companies
For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.



Example 1:

Flipped Trees Diagram
Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.
Example 2:

Input: root1 = [], root2 = []
Output: true
Example 3:

Input: root1 = [], root2 = [1]
Output: false


Constraints:

The number of nodes in each tree is in the range [0, 100].
Each tree will have unique node values in the range [0, 99].

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