给定一棵树，任务是找到给定树中每个节点到另一个节点最远的节点。
例子
Input:
Output: 2 3 3 3 4 4 4
Explanation:
Maximum Distance from Node 1 : 2 (Nodes {5, 6, 7} are at a distance 2)
Maximum Distance from Node 2 : 3 (Nodes {6, 7} are at a distance 3)
Maximum Distance from Node 3 : 3 (Nodes {5, 6, 7} are at a distance 3)
Maximum Distance from Node 4 : 3 (Node {5} is at a distance 3)
Maximum Distance from Node 5 : 4 (Nodes {6, 7} are at a distance 4)
Maximum Distance from Node 6 : 4 (Node {5} is at a distance 4)
Maximum Distance from Node 7 : 4 (Node {5} is at a distance 4)
Input:
Output : 3 2 3 3 2 3
方法：
请按照以下步骤解决问题：
- 使用 DFS 计算树的每个节点的高度(假设叶节点的高度为 1)
- 这给出了从节点到其子树中存在的所有节点的最大距离。存储这些高度。
- 现在，执行DFS以计算节点与其所有祖先的最大距离。存储这些距离。
- 对于每个节点，打印计算出的两个距离中的最大值。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
#define maxN 100001
// Adjacency List to store the graph
vector adj[maxN];
// Stores the height of each node
int height[maxN];
// Stores the maximum distance of a
// node from its ancestors
int dist[maxN];
// Function to add edge between
// two vertices
void addEdge(int u, int v)
{
// Insert edge from u to v
adj[u].push_back(v);
// Insert edge from v to u
adj[v].push_back(u);
}
// Function to calculate height of
// each Node
void dfs1(int cur, int par)
{
// Iterate in the adjacency
// list of the current node
for (auto u : adj[cur]) {
if (u != par) {
// Dfs for child node
dfs1(u, cur);
// Calculate height of nodes
height[cur]
= max(height[cur], height[u]);
}
}
// Increase height
height[cur] += 1;
}
// Function to calculate the maximum
// distance of a node from its ancestor
void dfs2(int cur, int par)
{
int max1 = 0;
int max2 = 0;
// Iterate in the adjacency
// list of the current node
for (auto u : adj[cur]) {
if (u != par) {
// Find two children
// with maximum heights
if (height[u] >= max1) {
max2 = max1;
max1 = height[u];
}
else if (height[u] > max2) {
max2 = height[u];
}
}
}
int sum = 0;
for (auto u : adj[cur]) {
if (u != par) {
// Calculate the maximum distance
// with ancestor for every node
sum = ((max1 == height[u]) ? max2 : max1);
if (max1 == height[u])
dist[u]
= 1 + max(1 + max2, dist[cur]);
else
dist[u]
= 1 + max(1 + max1, dist[cur]);
// Calculating for children
dfs2(u, cur);
}
}
}
// Driver Code
int main()
{
int n = 6;
addEdge(1, 2);
addEdge(2, 3);
addEdge(2, 4);
addEdge(2, 5);
addEdge(5, 6);
// Calculate height of
// nodes of the tree
dfs1(1, 0);
// Calculate the maximum
// distance with ancestors
dfs2(1, 0);
// Print the maximum of the two
// distances from each node
for (int i = 1; i <= n; i++)
cout << (max(dist[i], height[i]) - 1) << " ";
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
static final int maxN = 100001;
// Adjacency List to store the graph
@SuppressWarnings("unchecked")
static Vector []adj = new Vector[maxN];
// Stores the height of each node
static int []height = new int[maxN];
// Stores the maximum distance of a
// node from its ancestors
static int []dist = new int[maxN];
// Function to add edge between
// two vertices
static void addEdge(int u, int v)
{
// Insert edge from u to v
adj[u].add(v);
// Insert edge from v to u
adj[v].add(u);
}
// Function to calculate height of
// each Node
static void dfs1(int cur, int par)
{
// Iterate in the adjacency
// list of the current node
for(int u : adj[cur])
{
if (u != par)
{
// Dfs for child node
dfs1(u, cur);
// Calculate height of nodes
height[cur] = Math.max(height[cur],
height[u]);
}
}
// Increase height
height[cur] += 1;
}
// Function to calculate the maximum
// distance of a node from its ancestor
static void dfs2(int cur, int par)
{
int max1 = 0;
int max2 = 0;
// Iterate in the adjacency
// list of the current node
for(int u : adj[cur])
{
if (u != par)
{
// Find two children
// with maximum heights
if (height[u] >= max1)
{
max2 = max1;
max1 = height[u];
}
else if (height[u] > max2)
{
max2 = height[u];
}
}
}
int sum = 0;
for(int u : adj[cur])
{
if (u != par)
{
// Calculate the maximum distance
// with ancestor for every node
sum = ((max1 == height[u]) ?
max2 : max1);
if (max1 == height[u])
dist[u] = 1 + Math.max(1 + max2,
dist[cur]);
else
dist[u] = 1 + Math.max(1 + max1,
dist[cur]);
// Calculating for children
dfs2(u, cur);
}
}
}
// Driver Code
public static void main(String[] args)
{
int n = 6;
for(int i = 0; i < adj.length; i++)
adj[i] = new Vector();
addEdge(1, 2);
addEdge(2, 3);
addEdge(2, 4);
addEdge(2, 5);
addEdge(5, 6);
// Calculate height of
// nodes of the tree
dfs1(1, 0);
// Calculate the maximum
// distance with ancestors
dfs2(1, 0);
// Print the maximum of the two
// distances from each node
for(int i = 1; i <= n; i++)
System.out.print((Math.max(dist[i],
height[i]) - 1) + " ");
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to implement
# the above approach
maxN = 100001
# Adjacency List to store the graph
adj = [[] for i in range(maxN)]
# Stores the height of each node
height = [0 for i in range(maxN)]
# Stores the maximum distance of a
# node from its ancestors
dist = [0 for i in range(maxN)]
# Function to add edge between
# two vertices
def addEdge(u, v):
# Insert edge from u to v
adj[u].append(v)
# Insert edge from v to u
adj[v].append(u)
# Function to calculate height of
# each Node
def dfs1(cur, par):
# Iterate in the adjacency
# list of the current node
for u in adj[cur]:
if (u != par):
# Dfs for child node
dfs1(u, cur)
# Calculate height of nodes
height[cur] = max(height[cur],
height[u])
# Increase height
height[cur] += 1
# Function to calculate the maximum
# distance of a node from its ancestor
def dfs2(cur, par):
max1 = 0
max2 = 0
# Iterate in the adjacency
# list of the current node
for u in adj[cur]:
if (u != par):
# Find two children
# with maximum heights
if (height[u] >= max1):
max2 = max1
max1 = height[u]
elif (height[u] > max2):
max2 = height[u]
sum = 0
for u in adj[cur]:
if (u != par):
# Calculate the maximum distance
# with ancestor for every node
sum = (max2 if (max1 == height[u]) else max1)
if (max1 == height[u]):
dist[u] = 1 + max(1 + max2, dist[cur])
else:
dist[u] = 1 + max(1 + max1, dist[cur])
# Calculating for children
dfs2(u, cur)
# Driver Code
if __name__=="__main__":
n = 6
addEdge(1, 2)
addEdge(2, 3)
addEdge(2, 4)
addEdge(2, 5)
addEdge(5, 6)
# Calculate height of
# nodes of the tree
dfs1(1, 0)
# Calculate the maximum
# distance with ancestors
dfs2(1, 0)
# Print the maximum of the two
# distances from each node
for i in range(1, n + 1):
print(max(dist[i],
height[i]) - 1, end = ' ')
# This code is contributed by rutvik_56C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
static readonly int maxN = 100001;
// Adjacency List to store the graph
static List []adj = new List[maxN];
// Stores the height of each node
static int []height = new int[maxN];
// Stores the maximum distance of a
// node from its ancestors
static int []dist = new int[maxN];
// Function to add edge between
// two vertices
static void addEdge(int u, int v)
{
// Insert edge from u to v
adj[u].Add(v);
// Insert edge from v to u
adj[v].Add(u);
}
// Function to calculate height of
// each Node
static void dfs1(int cur, int par)
{
// Iterate in the adjacency
// list of the current node
foreach(int u in adj[cur])
{
if (u != par)
{
// Dfs for child node
dfs1(u, cur);
// Calculate height of nodes
height[cur] = Math.Max(height[cur],
height[u]);
}
}
// Increase height
height[cur] += 1;
}
// Function to calculate the maximum
// distance of a node from its ancestor
static void dfs2(int cur, int par)
{
int max1 = 0;
int max2 = 0;
// Iterate in the adjacency
// list of the current node
foreach(int u in adj[cur])
{
if (u != par)
{
// Find two children
// with maximum heights
if (height[u] >= max1)
{
max2 = max1;
max1 = height[u];
}
else if (height[u] > max2)
{
max2 = height[u];
}
}
}
int sum = 0;
foreach(int u in adj[cur])
{
if (u != par)
{
// Calculate the maximum distance
// with ancestor for every node
sum = ((max1 == height[u]) ?
max2 : max1);
if (max1 == height[u])
dist[u] = 1 + Math.Max(1 + max2,
dist[cur]);
else
dist[u] = 1 + Math.Max(1 + max1,
dist[cur]);
// Calculating for children
dfs2(u, cur);
}
}
}
// Driver Code
public static void Main(String[] args)
{
int n = 6;
for(int i = 0; i < adj.Length; i++)
adj[i] = new List();
addEdge(1, 2);
addEdge(2, 3);
addEdge(2, 4);
addEdge(2, 5);
addEdge(5, 6);
// Calculate height of
// nodes of the tree
dfs1(1, 0);
// Calculate the maximum
// distance with ancestors
dfs2(1, 0);
// Print the maximum of the two
// distances from each node
for(int i = 1; i <= n; i++)
Console.Write((Math.Max(dist[i],
height[i]) - 1) + " ");
}
}
// This code is contributed by Rohit_ranjan 输出：
3 2 3 3 2 3
时间复杂度： O(V+E)
辅助空间： O(N)
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