> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/trees/cousins-in-binary-tree.md).

# Cousins in Binary Tree

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.

**Example 1:**<br>

![](https://assets.leetcode.com/uploads/2019/02/12/q1248-01.png)

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

**Example 2:**<br>

![](https://assets.leetcode.com/uploads/2019/02/12/q1248-02.png)

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

**Example 3:**

![](https://assets.leetcode.com/uploads/2019/02/13/q1248-03.png)

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

**Constraints:**

* The number of nodes in the tree will be between `2` and `100`.
* Each node has a unique integer value from `1` to `100`.

```java
class Solution {
    public boolean isCousins(TreeNode root, int x, int y) {
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        while (q.size() != 0) {
            boolean xFound = false, yFound = false;
            int size = q.size();
            while (size-- > 0) {
                TreeNode node = q.poll();
                if (node.val == x)
                    xFound = true;
                if (node.val == y)
                    yFound = true;
                if (node.left != null)
                    q.add(node.left);
                if (node.right != null)
                    q.add(node.right);
                // If x and y belong to same parents
                if (node.left != null && node.right != null) {
                    if ((node.left.val == x && node.right.val == y) 
                     || (node.left.val == y && node.right.val == x))
                        return false;
                }
            }
            if (xFound && yFound)
                return true;
        }
        return false;
    }
}
```
