HackMD
687.Longest Univalue Path
tags: Medium,Tree,DFS
題目描述
Given the root of a binary tree, return the length of the longest path, where each node in the path has the same value. This path may or may not pass through the root.
The length of the path between two nodes is represented by the number of edges between them.
Example 1:
Image Not Showing
Possible Reasons
- The image file may be corrupted
- The server hosting the image is unavailable
- The image path is incorrect
- The image format is not supported
Input: root = [5,4,5,1,1,null,5]
Output: 2
Explanation: The shown image shows that the longest path of the same value (i.e. 5).
Example 2:
Image Not Showing
Possible Reasons
- The image file may be corrupted
- The server hosting the image is unavailable
- The image path is incorrect
- The image format is not supported
Input: root = [1,4,5,4,4,null,5]
Output: 2
Explanation: The shown image shows that the longest path of the same value (i.e. 4).
Constraints:
- The number of nodes in the tree is in the range [0, 104].
- -1000 <= Node.val <= 1000
- The depth of the tree will not exceed 1000.
解答
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 __init__(self):
self.mx = 0
def longestUnivaluePath(self, root: Optional[TreeNode]) -> int:
if not root: return 0
def dfs(root: Optional[TreeNode], num: int) -> int:
if not root: return 0
l = dfs(root.left, root.val)
r = dfs(root.right, root.val)
self.mx = l + r if l + r > self.mx else self.mx
return 0 if root.val != num else 1 + max(l, r)
dfs(root, root.val)
return self.mx
Kobe BryantNov 25, 2022
