"""
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:
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

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

# 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: TreeNode, x: int, y: int) -> bool:
        def check(node, mom, x, level):
            if not node:
                return
            if node.val == x:
                print (mom, level)
                return [mom, level]
            return check(node.left, node, x, level+1) or check(node.right, node, x, level+1)
        
        i = check(root, None, x, 0)
        j = check(root, None, y, 0)
        if i[0] != j[0] and i[1]==j[1]:
            return True
        return False

"""
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:
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

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

# 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: TreeNode, x: int, y: int) -> bool:
        def check(node, mom, x, level):
            if not node:
                return
            if node.val == x:
                print (mom, level)
                return [mom, level]
            return check(node.left, node, x, level+1) or check(node.right, node, x, level+1)
        
        i = check(root, None, x, 0)
        j = check(root, None, y, 0)
        if i[0] != j[0] and i[1]==j[1]:
            return True
        return False

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