Day-7_Cousins_in_Binary_Tree.py

'''
    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:
        bfs = deque([(root, None)])
        while bfs:
            n = len(bfs)
            parent = None
            foundone = False
            for _ in range(n):
                curr, par = bfs.popleft()
                if curr.val==x or curr.val==y:
                    if not foundone:
                        foundone = True
                        parent = par
                    else: return parent!=par
                if curr.right:
                    bfs.append((curr.right, curr))
                if curr.left:
                    bfs.append((curr.left, curr))
        return False