Problem Description
Given a tree, find the diameter of the tree, that is, the distance between the two furthest points on the tree.
Input description: the
number of input nodes, the two nodes of each edge, the weight of the edge
Output description:
output the longest path
Example
Example 1
Enter
6,[[0,1],[1,5],[1,2],[2,3],[2,4]],[3,4,2,1,5]
Output
11
Solutions
analysis
This question is mainly through depth-first traversal, and the maximum value is obtained by comparing and judging the weight value of each path.
method
- Using depth-first traversal, traverse the tree to get the maximum value
Code
public class Solution {
/**
* 树的直径
*
* @param n int整型 树的节点个数
* @param Tree_edge Interval类一维数组 树的边
* @param Edge_value int整型一维数组 边的权值
* @return int整型
*/
//6,[[0,1],[1,5],[1,2],[2,3],[2,4]],[3,4,2,1,5]
int res = 0;
public int solve(int n, Interval[] Tree_edge, int[] Edge_value) {
// write code here
// list数组存储每个节点的所有边
List<int[]>[] map = new ArrayList[n];
// 将所有的节点初始化
for (int i = 0; i < n; i++) map[i] = new ArrayList();
// 遍历所有的边，一次添加到对应数组的list中
for (int i = 0; i < Tree_edge.length; i++) {
// 起始节点的边，存储终节点和边的权重值
map[Tree_edge[i].start].add(new int[]{
Tree_edge[i].end, Edge_value[i]});
// 终节点的边，存储起节点和权重值
map[Tree_edge[i].end].add(new int[]{
Tree_edge[i].start, Edge_value[i]});
}
// 深度优先遍历，通过数组的bool值判断节点是否已经访问过
dfs(map, 0, new boolean[n]);
return res;
}
private int dfs(List<int[]>[] map, int index, boolean[] visited) {
// 将节点置为已经访问
visited[index] = true;
// 获取节点的所有的边
List<int[]> list = map[index];
int weight1 = 0, weight2 = 0;
// 遍历所有的边
for (int[] child2weight : list) {
// 获取边的另一端
int child = child2weight[0];
// 获取边的权重
int weight = child2weight[1];
// 判断边是否已经访问
if (visited[child]) continue;
// 计算路径长度
int num = weight + dfs(map, child, visited);
if (num > weight1) {
weight2 = weight1;
weight1 = num;
} else if (num > weight2) {
weight2 = num;
}
if (weight1 + weight2 > res) res = weight1 + weight2;
}
return Math.max(weight1, weight2);
}
}
If you want to test, you can go directly to the link of Niuke.com to do the test