LeetCode 687. Longest Univalue Path (dfs+bfs)

最新推荐文章于 2025-05-19 01:02:14 发布

原创最新推荐文章于 2025-05-19 01:02:14 发布 · 152 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 同时被 3 个专栏收录

18 篇文章

订阅专栏

算法

14 篇文章

订阅专栏

树

5 篇文章

订阅专栏

本文探讨了在二叉树中寻找最长相同值路径的问题，通过优化算法，使用一次深度优先搜索(DFS)替代广度优先搜索(BFS)，显著提高了运行效率。详细介绍了算法思路与实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a binary tree, find 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.

Note: The length of path between two nodes is represented by the number of edges between them.

Example 1:

Input:

Output:

Example 2:

Input:

Output:

思路
bfs遍历树，对任意一个节点，使用dfs，找到左侧与该节点相同的最长路径，在找到右侧与该节点相同的最长路径。使用dict记录值，减少寻找。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def dfs(self, root, parent):
        if root == None:
            return 0
        if parent != root.val:
            return 0
        tmp = self.l.get(root)
        tmp2 = self.r.get(root)
        if tmp!=None and tmp2 != None:
            return max(tmp, tmp2)
        left = self.dfs(root.left, root.val)
        right = self.dfs(root.right, root.val)
        self.l[root] = left
        self.r[root] = right
        return 1+max(left, right)
    
    def bfs(self, root):
        que = []
        que.insert(0, root)
        ans = 0
        while len(que) != 0:
            cur = que.pop()
            self.dfs(cur, cur.val)
            ans = max(ans, self.l[cur]+self.r[cur])
            if cur.left:
                que.insert(0, cur.left)
            if cur.right:
                que.insert(0, cur.right)
        return ans
    
    def longestUnivaluePath(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root == None: 
            return 0
        self.l = {}
        self.r = {}
        return self.bfs(root)

代码虽然AC了，但是运行时间很慢，进行优化下，使用一遍dfs即可。
判断如果结点不与父节点值一致就从当前开始。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def dfs(self, root, parent):
        if root == None:
            return 0
    
        tmp = self.l.get(root)
        tmp2 = self.r.get(root)
        if tmp!=None and tmp2 != None:
            return max(tmp, tmp2)
        
        left = self.dfs(root.left, root.val)
        right = self.dfs(root.right, root.val)
        self.l[root] = left
        self.r[root] = right
        self.ans = max(self.ans, left+right)
        if parent != root.val:  
            return 0
        return 1+max(left, right)
    
    
    def longestUnivaluePath(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root == None: 
            return 0
        self.l = {}
        self.r = {}
        self.ans = 0
        self.dfs(root, root.val)
        return self.ans