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

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

Example 3:
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.
"""
from typing import Optional


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

class CousinsInBinaryTree:
    def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
        parent = {}
        depth = {}

        def dfs(cur, pre, d):
            if not cur:
                return
            depth[cur.val] = d
            parent[cur.val] = pre
            dfs(cur.left, cur, d + 1)
            dfs(cur.right, cur, d + 1)

        dfs(root, -1, 0)
        return depth[x] == depth[y] and parent[x] != parent[y]

