## DEV Community

Abhishek Chaudhary

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# Cousins in Binary Tree

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.

SOLUTION:

``````# 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 inorder(self, root, l, x, y, path):
if root:
if root.left:
self.inorder(root.left, l + 1, x, y, path[-1:] + [root.left.val])
if root.val == x:
self.xnode = (l, path[-2] if len(path) >= 2 else None)
if root.val == y:
self.ynode = (l, path[-2] if len(path) >= 2 else None)
if root.right:
self.inorder(root.right, l + 1, x, y, path[-1:] + [root.right.val])

def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
self.inorder(root, 0, x, y, [root.val])
return self.xnode[0] == self.ynode[0] and self.xnode[1] != self.ynode[1]
``````