Mu solution to [Leetcode problem Cousins in Binary Tree](https://leetcode.com/problems/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;
    }
};

```