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Mu solution to Leetcode problem Cousins in Binary Tree works, but the code feels bulky. Is there a better way to solve this problem, particularly to use less additional variables? I will appreciate any suggestions on how to improve this code.

Problem:

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

My code:

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
private:
 void is_cousins(TreeNode* root, int to_find, int depth, 
 TreeNode* parent, pair<int, TreeNode*>& pair) {
 
 if (!root)
 return;
 
 is_cousins(root->left, to_find, depth + 1, root, pair);
 is_cousins(root->right, to_find, depth + 1, root, pair);
 
 if (root->val == to_find) {
 pair.first = depth;
 pair.second = parent;
 return;
 }
}
public:
 bool isCousins(TreeNode* root, int x, int y) {
 
 pair<int, TreeNode*> pairx = make_pair(0, nullptr);
 pair<int, TreeNode*> pairy = make_pair(0, nullptr);
 
 is_cousins(root, x, 0, nullptr, pairx);
 is_cousins(root, y, 0, nullptr, pairy);
 if (pairx.first == pairy.first && pairx.second != pairy.second)
 return true;
 return false;
 }
};