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

<p>Two nodes of a binary tree are <strong>cousins</strong> if they have the same depth with different parents.</p>

<p>Note that in a binary tree, the root node is at the depth <code>0</code>, and children of each depth <code>k</code> node are at the depth <code>k + 1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/12/q1248-01.png" style="width: 304px; height: 270px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4], x = 4, y = 3
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/12/q1248-02.png" style="width: 334px; height: 266px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,4,null,5], x = 5, y = 4
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/13/q1248-03.png" style="width: 267px; height: 258px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,4], x = 2, y = 3
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 100]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 100</code></li>
	<li>Each node has a <strong>unique</strong> value.</li>
	<li><code>x != y</code></li>
	<li><code>x</code> and <code>y</code> are exist in the tree.</li>
</ul>

from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def isCousins(self, root: TreeNode, x: int, y: int) -> bool:
        if not root:
            return False
        
        # Perform BFS to find nodes x and y
        queue = deque([(root, None, 0)])  # (node, parent, depth)
        x_info, y_info = None, None
        
        while queue:
            node, parent, depth = queue.popleft()
            
            if node.val == x:
                x_info = (parent, depth)
            if node.val == y:
                y_info = (parent, depth)
            
            if x_info and y_info:
                break
            
            if node.left:
                queue.append((node.left, node, depth + 1))
            if node.right:
                queue.append((node.right, node, depth + 1))
        
        # Check if both nodes are found and are cousins
        if x_info and y_info:
            x_parent, x_depth = x_info
            y_parent, y_depth = y_info
            return x_depth == y_depth and x_parent != y_parent
        
        return False

# Example usage:
# Create the binary tree for Example 2
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.right = TreeNode(4)
root.right.right = TreeNode(5)

solution = Solution()
print(solution.isCousins(root, 5, 4))  # Output: True
16300