<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>

# Definition for a binary tree node.
# 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: Optional[TreeNode], x: int, y: int) -> bool:
        def dfs(node, parent, depth, target, result):
            if not node:
                return False
            if node.val == target:
                result[0], result[1] = parent, depth
                return True
            return dfs(node.left, node, depth + 1, target, result) or dfs(node.right, node, depth + 1, target, result)

        # To store the parent and depth of nodes x and y
        x_info, y_info = [None, 0], [None, 0]

        # Perform DFS to find x and y
        dfs(root, None, 0, x, x_info)
        dfs(root, None, 0, y, y_info)

        # x and y are cousins if their depth is the same but they have different parents
        return x_info[0] != y_info[0] and x_info[1] == y_info[1]
        
16500