给定一棵N叉树，任务是找到给定树的最大宽度。树的最大宽度是所有级别中宽度的最大值。
例子：
Input:
Output: 4
Explanation:
Width of 0th level is 1.
Width of 1st level is 3.
Width of 2nd level is 4.
Therefore, the maximum width is 4
Input:
Output:4
方法：这个问题可以用BFS解决。这个想法是执行树的级别顺序遍历。在做遍历的时候，分别处理不同层次的节点。对于正在处理的每个级别，计算每个级别存在的节点数并跟踪最大计数。请按照以下步骤解决问题：
- 初始化一个变量，比如maxWidth来存储树所需的最大宽度。
- 初始化队列以执行给定树的级别顺序遍历。
- 将根节点推入队列。
- 如果队列的大小超过maxWidth任何级别，然后更新maxWidth到 队列的大小。
- 遍历队列，将下一层的所有节点推入队列，弹出当前层的所有节点。
- 重复上述步骤，直到遍历树的所有级别。
- 最后，返回maxWidth的最终值。
下面是上述方法的实现：
C++
4
/ | \
2 3 -5
/ \ /\
-1 3 -2 6Java
1
/ | \
2 -1 3
/ \ \
4 5 8
/ / | \
2 6 12 7 Python3
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the maximum width of
// the tree using level order traversal
int maxWidth(int N, int M,
vector cost,
vector > s)
{
// Store the edges of the tree
vector adj[N];
for (int i = 0; i < M; i++) {
adj[s[i][0]].push_back(
s[i][1]);
}
// Stores maximum width
// of the tree
int result = 0;
// Stores the nodes
// of each level
queue q;
// Insert root node
q.push(0);
// Perform level order
// traversal on the tree
while (!q.empty()) {
// Stores the size of
// the queue
int count = q.size();
// Update maximum width
result = max(count, result);
// Push the nodes of the next
// level and pop the elements
// of the current level
while (count--) {
// Get element from the
// front the Queue
int temp = q.front();
q.pop();
// Push all nodes of the next level.
for (int i = 0; i < adj[temp].size();
i++) {
q.push(adj[temp][i]);
}
}
}
// Return the result.
return result;
}
// Driver Code
int main()
{
int N = 11, M = 10;
vector > edges;
edges.push_back({ 0, 1 });
edges.push_back({ 0, 2 });
edges.push_back({ 0, 3 });
edges.push_back({ 1, 4 });
edges.push_back({ 1, 5 });
edges.push_back({ 3, 6 });
edges.push_back({ 4, 7 });
edges.push_back({ 6, 10 });
edges.push_back({ 6, 8 });
edges.push_back({ 6, 9 });
vector cost
= { 1, 2, -1, 3, 4, 5,
8, 2, 6, 12, 7 };
/* Constructed tree is:
1
/ | \
2 -1 3
/ \ \
4 5 8
/ / | \
2 6 12 7
*/
cout << maxWidth(N, M, cost, edges);
return 0;
} C#
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to find the maximum width of
// the tree using level order traversal
static int maxWidth(int N, int M,ArrayList cost,
ArrayList > s)
{
// Store the edges of the tree
ArrayList > adj =
new ArrayList >();
for(int i = 0; i < N; i++)
{
adj.add(new ArrayList());
}
for(int i = 0; i < M; i++)
{
adj.get(s.get(i).get(0)).add(s.get(i).get(1));
}
// Stores maximum width
// of the tree
int result = 0;
// Stores the nodes
// of each level
Queue q = new LinkedList<>();
// Insert root node
q.add(0);
// Perform level order
// traversal on the tree
while(q.size() != 0)
{
// Stores the size of
// the queue
int count = q.size();
// Update maximum width
result = Math.max(count, result);
// Push the nodes of the next
// level and pop the elements
// of the current level
while(count-->0)
{
// Get element from the
// front the Queue
int temp = q.remove();
// Push all nodes of the next level.
for(int i = 0; i < adj.get(temp).size(); i++)
{
q.add(adj.get(temp).get(i));
}
}
}
// Return the result.
return result;
}
// Driver Code
public static void main (String[] args)
{
int N = 11, M = 10;
ArrayList > edges = new ArrayList >();
edges.add(new ArrayList(Arrays.asList( 0, 1)));
edges.add(new ArrayList(Arrays.asList( 0, 2)));
edges.add(new ArrayList(Arrays.asList( 0, 3)));
edges.add(new ArrayList(Arrays.asList(1,4)));
edges.add(new ArrayList(Arrays.asList(1,5)));
edges.add(new ArrayList(Arrays.asList(3,6)));
edges.add(new ArrayList(Arrays.asList(4,7)));
edges.add(new ArrayList(Arrays.asList(6,10)));
edges.add(new ArrayList(Arrays.asList(6,8)));
edges.add(new ArrayList(Arrays.asList(6,9)));
ArrayList cost = new ArrayList(Arrays.asList(1, 2, -1, 3, 4, 5,8, 2, 6, 12, 7 ));
/* Constructed tree is:
1
/ | \
2 -1 3
/ \ \
4 5 8
/ / | \
2 6 12 7
*/
System.out.println(maxWidth(N, M, cost, edges));
}
}
// This code is contributed by avanitrachhadiya2155 输出：
# Python3 program to implement
# the above approach
from collections import deque
# Function to find the maximum width of
#. he tree using level order traversal
def maxWidth(N, M, cost, s):
# Store the edges of the tree
adj = [[] for i in range(N)]
for i in range(M):
adj[s[i][0]].append(s[i][1])
# Stores maximum width
# of the tree
result = 0
# Stores the nodes
# of each level
q = deque()
# Insert root node
q.append(0)
# Perform level order
# traversal on the tree
while (len(q) > 0):
# Stores the size of
# the queue
count = len(q)
# Update maximum width
result = max(count, result)
# Push the nodes of the next
# level and pop the elements
# of the current level
while (count > 0):
# Get element from the
# front the Queue
temp = q.popleft()
# Push all nodes of the next level.
for i in adj[temp]:
q.append(i)
count -= 1
# Return the result.
return result
# Driver Code
if __name__ == '__main__':
N = 11
M = 10
edges = []
edges.append([0, 1])
edges.append([0, 2])
edges.append([0, 3])
edges.append([1, 4])
edges.append([1, 5])
edges.append([3, 6])
edges.append([4, 7])
edges.append([6, 1])
edges.append([6, 8])
edges.append([6, 9])
cost = [ 1, 2, -1, 3, 4, 5,
8, 2, 6, 12, 7]
# Constructed tree is:
# 1
# / | \
# 2 -1 3
# / \ \
# 4 5 8
# / / | \
# 2 6 12 7
print(maxWidth(N, M, cost, edges))
# This code is contributed by mohit kumar 29时间复杂度： O(N)
辅助空间： O(N)
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