# Cousins In Binary Tree Problem

## Description

LeetCode Problem 993.

Given the root of a binary tree with unique values and the values of two different nodes of the tree x and y, return true if the nodes corresponding to the values x and y in the tree are cousins, or false otherwise.

Two nodes of a binary tree are cousins if they have the same depth with different parents. Note that in a binary tree, the root node is at the depth 0, and children of each depth k node are at the depth k + 1.

Example 1:

``````1
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Input: root = [1,2,3,4], x = 4, y = 3
Output: false
``````

Example 2:

``````1
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Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true
``````

Example 3:

``````1
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Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false
``````

Constraints:

• The number of nodes in the tree is in the range [2, 100].
• 1 <= Node.val <= 100
• Each node has a unique value.
• x != y
• x and y are exist in the tree.

## Sample C++ Code

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/**
* 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 {
public:
void rec(TreeNode* root, int x, int y, int d, int parent) {
if (!root) return;

if (root->val == x)
dx = d, x_par = parent;
if (root->val == y)
dy = d, y_par = parent;

if (dx != -1 && dy != -1) return;

rec(root->left, x, y, d+1, root->val);
rec(root->right, x, y, d+1, root->val);
}

bool isCousins(TreeNode* root, int x, int y) {
rec(root, x, y, 0, -1);
return dx == dy && x_par != y_par;
}

private:
int dx = -1, dy = -1, x_par = -1, y_par = -1;
};
``````