Description
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.
分析
题目的意思是:找出二叉树的堂兄弟结点,两个结点在同一层并且没有共同的直接父结点。用递归的话就是找出两个结点所在的层,并且记录其父结点。最后判断一下就行了。
代码
# 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 solve(self,root,x,y,d,parent):
if(root is None):
return
if(root.val==x):
self.a=d
self.aNode=parent
elif(root.val==y):
self.b=d
self.bNode=parent
self.solve(root.left,x,y,d+1,root)
self.solve(root.right,x,y,d+1,root)
def isCousins(self, root: TreeNode, x: int, y: int) -> bool:
self.solve(root,x,y,0,None)
if(self.a==self.b and self.aNode!=self.bNode):
return True
return False
参考文献
[LeetCode] Python beats 98% (easy to understand) with explanation