Leetcode #993: Cousins in Binary Tree
In this guide, we solve Leetcode #993 Cousins in Binary Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root = [1,2,3,4], x = 4, y = 3
Output: false
Python 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 isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
q = deque([(root, None)])
depth = 0
p1 = p2 = None
d1 = d2 = None
while q:
for _ in range(len(q)):
node, parent = q.popleft()
if node.val == x:
p1, d1 = parent, depth
elif node.val == y:
p2, d2 = parent, depth
if node.left:
q.append((node.left, node))
if node.right:
q.append((node.right, node))
depth += 1
return p1 != p2 and d1 == d2
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.