树的直径求解方法

1 介绍

树的直径是指树中任意两个节点之间的最长路径长度。通常可以通过两次广度优先搜索（BFS）或深度优先搜索（DFS）来求树的直径：

第一次从任意一点出发，找到离它最远的节点 A。
第二次从节点 A 出发，找到离它最远的节点 B，A 和 B 之间的路径长度就是树的直径。

2 训练

解题思路：分别求出两棵树的直径，然后比较大小max(d1, d2, r1 + r2 + 1)即可。

python3代码如下，

py 复制代码

from collections import deque

class Solution:
    def minimumDiameterAfterMerge(self, edges1: List[List[int]], edges2: List[List[int]]) -> int:

        def bfs(graph, start):
            n = len(graph)
            dist = [-1] * n
            queue = deque([start])
            dist[start] = 0
            while queue:
                node = queue.popleft()
                for neighbor in graph[node]:
                    if dist[neighbor] == -1:
                        dist[neighbor] = dist[node] + 1
                        queue.append(neighbor)
            farthest_node = dist.index(max(dist))
            return farthest_node, dist

        def find_diameter_and_radius(graph):
            node1, _ = bfs(graph, 0)
            node2, dist = bfs(graph, node1)
            diameter = dist[node2]
            return diameter, dist

        n = len(edges1) + 1 
        m = len(edges2) + 1
        graph1 = [[] for _ in range(n)]
        graph2 = [[] for _ in range(m)]
        
        for u, v in edges1:
            graph1[u].append(v)
            graph1[v].append(u)
            
        for u, v in edges2:
            graph2[u].append(v)
            graph2[v].append(u)
            
        diameter1, dist1 = find_diameter_and_radius(graph1)
        diameter2, dist2 = find_diameter_and_radius(graph2)
        
        # Finding radius of each tree
        radius1 = (diameter1 + 1) // 2
        radius2 = (diameter2 + 1) // 2
        
        # Minimum new diameter
        min_diameter = max(diameter1, diameter2, radius1 + radius2 + 1)
        
        return min_diameter

树的直径求解方法

目录

1 介绍

2 训练