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 -
xandyare 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]
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