Cousins in Binary Tree

#leetcode #dsa #theabbie

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]