打印其中包含奇数和偶数节点的所有级别 |第 2 组

给定一棵N叉树，打印其中包含奇数和偶数节点的所有级别。

例子：

For example consider the following tree
          1               - Level 1
       /     \
      2       3           - Level 2
    /   \       \
   4     5       6        - Level 3
        /  \     /
       7    8   9         - Level 4

The levels with odd number of nodes are: 1 3 4 
The levels with even number of nodes are: 2

注意：级别编号从 1 开始，即根节点在级别 1。

方法：

• 将所有连接节点插入二维向量树。
• 在树上运行 BFS，使得 height[node] = 1 + height[parent]
• 一旦 BFS 遍历完成，对于每个节点的级别，将 count[] 数组增加 1。
• 从第一层迭代到最后一层，并将所有具有 count[] 值的节点打印为奇数，以获得奇数节点的级别。
• 从第一级迭代到最后一级，并将所有具有 count[] 值的节点打印为偶数，以获得偶数节点的级别。

下面是上述方法的实现：

C++

// C++ program to print all levels
// with odd and even number of nodes
 
#include 
using namespace std;
 
// Function for BFS in a tree
void bfs(int node, int parent, int height[], int vis[],
         vector tree[])
{
 
    // mark first node as visited
    vis[node] = 1;
 
    // Declare Queue
    queue q;
 
    // Push the first element
    q.push(1);
 
    // calculate the level of every node
    height[node] = 1 + height[parent];
 
    // Check if the queue is empty or not
    while (!q.empty()) {
 
        // Get the top element in the queue
        int top = q.front();
 
        // pop the element
        q.pop();
 
        // mark as visited
        vis[top] = 1;
 
        // Iterate for the connected nodes
        for (int i = 0; i < tree[top].size(); i++) {
 
            // if not visited
            if (!vis[tree[top][i]]) {
 
                // Insert into queue
                q.push(tree[top][i]);
 
                // Increase level
                height[tree[top][i]] = 1 + height[top];
            }
        }
    }
}
 
// Function to insert edges
void insertEdges(int x, int y, vector tree[])
{
    tree[x].push_back(y);
    tree[y].push_back(x);
}
 
// Function to print all levels
void printLevelsOddEven(int N, int vis[], int height[])
{
    int mark[N + 1];
    memset(mark, 0, sizeof mark);
 
    int maxLevel = 0;
    for (int i = 1; i <= N; i++) {
 
        // count number of nodes
        // in every level
        if (vis[i])
            mark[height[i]]++;
 
        // find the maximum height of tree
        maxLevel = max(height[i], maxLevel);
    }
 
    // print odd number of nodes
    cout << "The levels with odd number of nodes are: ";
    for (int i = 1; i <= maxLevel; i++) {
        if (mark[i] % 2)
            cout << i << " ";
    }
 
    // print even number of nodes
    cout << "\nThe levels with even number of nodes are: ";
    for (int i = 1; i <= maxLevel; i++) {
        if (mark[i] % 2 == 0)
            cout << i << " ";
    }
}
 
// Driver Code
int main()
{
    // Construct the tree
 
    /*   1
       /   \
      2     3
     / \     \
    4    5    6
        / \  /
       7   8 9  */
 
    const int N = 9;
 
    vector tree[N + 1];
 
    insertEdges(1, 2, tree);
    insertEdges(1, 3, tree);
    insertEdges(2, 4, tree);
    insertEdges(2, 5, tree);
    insertEdges(5, 7, tree);
    insertEdges(5, 8, tree);
    insertEdges(3, 6, tree);
    insertEdges(6, 9, tree);
 
    int height[N + 1];
    int vis[N + 1] = { 0 };
 
    // call the bfs function
    bfs(1, 0, height, vis, tree);
 
    // Function to print
    printLevelsOddEven(N, vis, height);
 
    return 0;
}

Java

// Java program to print all levels
// with odd and even number of nodes
import java.util.*;
public class Main
{
    // Function for BFS in a tree
    static void bfs(int node, int parent, int[] height, int[] vis, Vector> tree)
    {
        // Mark first node as visited
        vis[node] = 1;
       
        // Declare Queue
        Queue q = new LinkedList<>();
       
        // Push the first element
        q.add(1);
       
        // Calculate the level of every node
        height[node] = 1 + height[parent];
       
        // Check if the queue is empty or not
        while (q.size() != 0)
        {
              
            // Get the top element in the queue
            int top = (int)q.peek();
       
            // Pop the element
            q.remove();
       
            // Mark as visited
            vis[top] = 1;
       
            // Iterate for the connected nodes
            for(int i = 0; i < tree.get(top).size(); i++)
            {
                  
                // If not visited
                if (vis[(int)tree.get(top).get(i)] == 0)
                {
                      
                    // Insert into queue
                    q.add(tree.get(top).get(i));
       
                    // Increase level
                    height[(int)tree.get(top).get(i)] = 1 + height[top];
                }
            }
        }
    }
       
    // Function to insert edges
    static void insertEdges(int x, int y, Vector> tree)
    {
        tree.get(x).add(y);
        tree.get(y).add(x);
    }
       
    // Function to print all levels
    static void printLevelsOddEven(int N, int[] vis, int[] height)
    {
        int[] mark = new int[N + 1];
        for(int i = 0; i < N + 1; i++)
        {
            mark[i] = 0;
        }
       
        int maxLevel = 0;
        for(int i = 1; i <= N; i++)
        {
              
            // Count number of nodes
            // in every level
            if (vis[i]!=0)
                mark[height[i]]++;
       
            // Find the maximum height of tree
            maxLevel = Math.max(height[i], maxLevel);
        }
       
        // Print odd number of nodes
        System.out.print("The levels with odd " +
                      "number of nodes are: ");
          
        for(int i = 1; i <= maxLevel; i++)
        {
            if (mark[i] % 2 != 0)
            {
                System.out.print(i + " ");
            }
        }
       
        // print even number of nodes
        System.out.println();
        System.out.print("The levels with even " +
                      "number of nodes are: ");
          
        for(int i = 1; i <= maxLevel; i++)
        {
            if (mark[i] % 2 == 0)
            {
                System.out.print(i + " ");
            }
        }
    }
 
    public static void main(String[] args) {
        // Construct the tree
      
        /*   1
           /   \
          2     3
         / \     \
        4    5    6
            / \  /
           7   8 9  */
          
        int N = 9;
          
        Vector> tree = new Vector>();
          
        for(int i = 0; i < N + 1; i++)
        {
            tree.add(new Vector());
        }
          
        insertEdges(1, 2, tree);
        insertEdges(1, 3, tree);
        insertEdges(2, 4, tree);
        insertEdges(2, 5, tree);
        insertEdges(5, 7, tree);
        insertEdges(5, 8, tree);
        insertEdges(3, 6, tree);
        insertEdges(6, 9, tree);
          
        int[] height = new int[N + 1];
        int[] vis = new int[N + 1];
        for(int i = 0; i < N + 1; i++)
        {
            vis[i] = 0;
        }
          
        height[0] = 0;
          
        // Call the bfs function
        bfs(1, 0, height, vis, tree);
          
        // Function to print
        printLevelsOddEven(N, vis, height);
    }
}
 
// This code is contributed by divyeshrabadiya07.

Python3

# Python3 program to print all levels
# with odd and even number of nodes
 
# Function for BFS in a tree
def bfs(node, parent, height, vis, tree):
 
    # mark first node as visited
    vis[node] = 1
 
    # Declare Queue
    q = []
 
    # append the first element
    q.append(1)
 
    # calculate the level of every node
    height[node] = 1 + height[parent]
 
    # Check if the queue is empty or not
    while (len(q)):
 
        # Get the top element in
        # the queue
        top = q[0]
 
        # pop the element
        q.pop(0)
 
        # mark as visited
        vis[top] = 1
 
        # Iterate for the connected nodes
        for i in range(len(tree[top])):
             
            # if not visited
            if (not vis[tree[top][i]]):
 
                # Insert into queue
                q.append(tree[top][i])
 
                # Increase level
                height[tree[top][i]] = 1 + height[top]
 
# Function to insert edges
def insertEdges(x, y, tree):
 
    tree[x].append(y)
    tree[y].append(x)
     
# Function to print all levels
def printLevelsOddEven(N, vis, height):
 
    mark = [0] * (N + 1)
 
    maxLevel = 0
    for i in range(1, N + 1):
 
        # count number of nodes
        # in every level
        if (vis[i]) :
            mark[height[i]] += 1
 
        # find the maximum height of tree
        maxLevel = max(height[i], maxLevel)
     
    # print odd number of nodes
    print("The levels with odd number",
          "of nodes are:", end = " ")
    for i in range(1, maxLevel + 1):
        if (mark[i] % 2):
            print(i, end = " " )
     
    # print even number of nodes
    print("\nThe levels with even number",
            "of nodes are:", end = " ")
    for i in range(1, maxLevel ):
        if (mark[i] % 2 == 0):
            print(i, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    # Construct the tree
    """ 1
    / \
    2 3
    / \ \
    4 5 6
        / \ /
    7 8 9 """
 
    N = 9
 
    tree = [[0]] * (N + 1)
 
    insertEdges(1, 2, tree)
    insertEdges(1, 3, tree)
    insertEdges(2, 4, tree)
    insertEdges(2, 5, tree)
    insertEdges(5, 7, tree)
    insertEdges(5, 8, tree)
    insertEdges(3, 6, tree)
    insertEdges(6, 9, tree)
 
    height = [0] * (N + 1)
    vis = [0] * (N + 1)
 
    # call the bfs function
    bfs(1, 0, height, vis, tree)
 
    # Function to print
    printLevelsOddEven(N, vis, height)
 
# This code is contributed
# by SHUBHAMSINGH10

C#

// C# program to print all levels
// with odd and even number of nodes
using System;
using System.Collections;
 
class GFG{
 
// Function for BFS in a tree
static void bfs(int node, int parent,
                int []height, int []vis,
                ArrayList []tree)
{
     
    // Mark first node as visited
    vis[node] = 1;
  
    // Declare Queue
    Queue q = new Queue();
  
    // Push the first element
    q.Enqueue(1);
  
    // Calculate the level of every node
    height[node] = 1 + height[parent];
  
    // Check if the queue is empty or not
    while (q.Count != 0)
    {
         
        // Get the top element in the queue
        int top = (int)q.Peek();
  
        // Pop the element
        q.Dequeue();
  
        // Mark as visited
        vis[top] = 1;
  
        // Iterate for the connected nodes
        for(int i = 0; i < tree[top].Count; i++)
        {
             
            // If not visited
            if (vis[(int)tree[top][i]] == 0)
            {
                 
                // Insert into queue
                q.Enqueue(tree[top][i]);
  
                // Increase level
                height[(int)tree[top][i]] = 1 + height[top];
            }
        }
    }
}
  
// Function to insert edges
static void insertEdges(int x, int y, ArrayList []tree)
{
    tree[x].Add(y);
    tree[y].Add(x);
}
  
// Function to print all levels
static void printLevelsOddEven(int N, int []vis,
                               int []height)
{
    int []mark = new int[N + 1];
    Array.Fill(mark, 0);
  
    int maxLevel = 0;
    for(int i = 1; i <= N; i++)
    {
         
        // Count number of nodes
        // in every level
        if (vis[i]!=0)
            mark[height[i]]++;
  
        // Find the maximum height of tree
        maxLevel = Math.Max(height[i], maxLevel);
    }
  
    // Print odd number of nodes
    Console.Write("The levels with odd " +
                  "number of nodes are: ");
     
    for(int i = 1; i <= maxLevel; i++)
    {
        if (mark[i] % 2 != 0)
        {
            Console.Write(i + " ");
        }
    }
  
    // print even number of nodes
    Console.Write("\nThe levels with even " +
                  "number of nodes are: ");
     
    for(int i = 1; i <= maxLevel; i++)
    {
        if (mark[i] % 2 == 0)
        {
            Console.Write(i + " ");
        }
    }
}
     
// Driver code   
static void Main()
{
       
    // Construct the tree
     
    /*   1
       /   \
      2     3
     / \     \
    4    5    6
        / \  /
       7   8 9  */
     
    int N = 9;
     
    ArrayList []tree = new ArrayList[N + 1];
     
    for(int i = 0; i < N + 1; i++)
    {
        tree[i] = new ArrayList();
    }
     
    insertEdges(1, 2, tree);
    insertEdges(1, 3, tree);
    insertEdges(2, 4, tree);
    insertEdges(2, 5, tree);
    insertEdges(5, 7, tree);
    insertEdges(5, 8, tree);
    insertEdges(3, 6, tree);
    insertEdges(6, 9, tree);
     
    int []height = new int[N + 1];
    int []vis = new int[N + 1];
    Array.Fill(vis, 0);
     
    height[0] = 0;
     
    // Call the bfs function
    bfs(1, 0, height, vis, tree);
     
    // Function to print
    printLevelsOddEven(N, vis, height);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
    // JavaScript program to print all levels
    // with odd and even number of nodes
     
    // Function for BFS in a tree
    function bfs(node, parent, height, vis, tree)
    {
 
        // Mark first node as visited
        vis[node] = 1;
 
        // Declare Queue
        let q = [];
 
        // Push the first element
        q.push(1);
 
        // Calculate the level of every node
        height[node] = 1 + height[parent];
 
        // Check if the queue is empty or not
        while (q.length != 0)
        {
 
            // Get the top element in the queue
            let top = q[0];
 
            // Pop the element
            q.shift();
 
            // Mark as visited
            vis[top] = 1;
 
            // Iterate for the connected nodes
            for(let i = 0; i < tree[top].length; i++)
            {
 
                // If not visited
                if (vis[tree[top][i]] == 0)
                {
 
                    // Insert into queue
                    q.push(tree[top][i]);
 
                    // Increase level
                    height[tree[top][i]] = 1 + height[top];
                }
            }
        }
    }
 
    // Function to insert edges
    function insertEdges(x, y, tree)
    {
        tree[x].push(y);
        tree[y].push(x);
    }
 
    // Function to print all levels
    function printLevelsOddEven(N, vis, height)
    {
        let mark = new Array(N + 1);
        mark.fill(0);
 
        let maxLevel = 0;
        for(let i = 1; i <= N; i++)
        {
 
            // Count number of nodes
            // in every level
            if (vis[i]!=0)
                mark[height[i]]++;
 
            // Find the maximum height of tree
            maxLevel = Math.max(height[i], maxLevel);
        }
 
        // Print odd number of nodes
        document.write("The levels with odd " +
                      "number of nodes are: ");
 
        for(let i = 1; i <= maxLevel; i++)
        {
            if (mark[i] % 2 != 0)
            {
                document.write(i + " ");
            }
        }
 
        // print even number of nodes
           document.write("</br>" + "The levels with even " +
                      "number of nodes are: ");
 
        for(let i = 1; i <= maxLevel; i++)
        {
            if (mark[i] % 2 == 0)
            {
                document.write(i + " ");
            }
        }
    }
     
    // Construct the tree
      
    /*   1
       /   \
      2     3
     / \     \
    4    5    6
        / \  /
       7   8 9  */
      
    let N = 9;
      
    let tree = new Array(N + 1);
      
    for(let i = 0; i < N + 1; i++)
    {
        tree[i] = [];
    }
      
    insertEdges(1, 2, tree);
    insertEdges(1, 3, tree);
    insertEdges(2, 4, tree);
    insertEdges(2, 5, tree);
    insertEdges(5, 7, tree);
    insertEdges(5, 8, tree);
    insertEdges(3, 6, tree);
    insertEdges(6, 9, tree);
      
    let height = new Array(N + 1);
    let vis = new Array(N + 1);
    vis.fill(0);
      
    height[0] = 0;
      
    // Call the bfs function
    bfs(1, 0, height, vis, tree);
      
    // Function to print
    printLevelsOddEven(N, vis, height);
 

输出：

The levels with odd number of nodes are: 1 3 4 
The levels with even number of nodes are: 2

时间复杂度：O(N)
辅助空间：O(N)