993. Cousins in Binary Tree-二叉树的堂兄弟节点

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

Constraints:

• The number of nodes in the tree will be between 2 and 100.
• Each node has a unique integer value from 1 to 100.

Solution in python:

# 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:
        flag = False
        temp = [root]
        level = []
        while temp:
            alist = [q.val for q in temp]
            if x in alist and y in alist:
                flag = True
            for item in temp:
                if item.left:
                    level.append(item.left)
                if item.right:
                    level.append(item.right)
            temp = level
            level = []
        flag2 = True
        def find(root, x, y):
            if root == None:
                return
            elif root.left and root.right == None:
                find(root.left, x, y)
            elif root.right and root.left == None:
                find(root.right, x, y)
            elif root.left and root.right:
                if (root.left.val == x and root.right.val == y) or (root.left.val == y and root.right.val == x):
                    nonlocal flag2
                    flag2 = False
                else:
                    find(root.left, x, y)
                    find(root.right, x, y)
            else:
                return

        find(root, x, y)
        return flag and flag2