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📜  按给定二叉树的级别顺序打印偶数个偶数级别的节点

📅  最后修改于: 2021-04-27 19:10:59             🧑  作者: Mango

给定一棵二叉树,任务是在树的层级遍历中打印奇数级的偶数定位节点。根被认为是在级别0处,而任何级别的最左边的节点都被认为是在位置0处的节点。

例子:

Input:
           1
         /    \
        2       3
      / \      /  \
     4   5    6    7
        /  \     
       8    9
      /      \
     10       11
Output: 2 8

Input:
      2
    /   \
   4     15
  /     /
 45   17
Output: 4

先决条件–均匀定位的元素处于均匀级别

方法:要逐级打印节点,请使用级顺序遍历。这个想法是基于逐行遍历打印级别顺序的。为此,逐级遍历节点,并在每级之后切换奇数级标志。同样,将每个级别的第一个节点标记为偶数位置,并在每次处理下一个节点后将其切换。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
  
struct Node {
    int data;
    Node *left, *right;
};
  
// Iterative method to do level order
// traversal line by line
void printOddLevelEvenNodes(Node* root)
{
    // Base Case
    if (root == NULL)
        return;
  
    // Create an empty queue for level
    // order traversal
    queue q;
  
    // Enqueue root and initialize level as even
    q.push(root);
    bool evenLevel = true;
  
    while (1) {
  
        // nodeCount (queue size) indicates
        // number of nodes in the current level
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
  
        // Mark 1st node as even positioned
        bool evenNodePosition = true;
  
        // Dequeue all the nodes of current level
        // and Enqueue all the nodes of next level
        while (nodeCount > 0) {
            Node* node = q.front();
  
            // Print only even positioned
            // nodes of even levels
            if (!evenLevel && evenNodePosition)
                cout << node->data << " ";
            q.pop();
            if (node->left != NULL)
                q.push(node->left);
            if (node->right != NULL)
                q.push(node->right);
            nodeCount--;
  
            // Switch the even position flag
            evenNodePosition = !evenNodePosition;
        }
  
        // Switch the even level flag
        evenLevel = !evenLevel;
    }
}
  
// Utility method to create a node
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
  
// Driver code
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->left->right->left = newNode(8);
    root->left->right->right = newNode(9);
    root->left->right->left->left = newNode(10);
    root->left->right->right->right = newNode(11);
  
    printOddLevelEvenNodes(root);
  
    return 0;
}


Java
// Java implementation of the approach
import java.util.*;
class GFG 
{
  
static class Node
{
    int data;
    Node left, right;
};
  
// Iterative method to do level order
// traversal line by line
static void printOddLevelEvenNodes(Node root)
{
    // Base Case
    if (root == null)
        return;
  
    // Create an empty queue for level
    // order traversal
    Queue q = new LinkedList<>();
  
    // Enqueue root and initialize level as even
    q.add(root);
    boolean evenLevel = true;
  
    while (true) 
    {
  
        // nodeCount (queue size) indicates
        // number of nodes in the current level
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
  
        // Mark 1st node as even positioned
        boolean evenNodePosition = true;
  
        // Dequeue all the nodes of current level
        // and Enqueue all the nodes of next level
        while (nodeCount > 0) 
        {
            Node node = q.peek();
  
            // Print only even positioned
            // nodes of even levels
            if (!evenLevel && evenNodePosition)
                System.out.print(node.data + " ");
            q.remove();
            if (node.left != null)
                q.add(node.left);
            if (node.right != null)
                q.add(node.right);
            nodeCount--;
  
            // Switch the even position flag
            evenNodePosition = !evenNodePosition;
        }
  
        // Switch the even level flag
        evenLevel = !evenLevel;
    }
}
  
// Utility method to create a node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
  
// Driver code
public static void main(String[] args) 
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.right.left = newNode(8);
    root.left.right.right = newNode(9);
    root.left.right.left.left = newNode(10);
    root.left.right.right.right = newNode(11);
  
    printOddLevelEvenNodes(root);
}
}
  
// This code is contributed by 29AjayKumar


Python3
# Python3 implementation of the approach 
  
# Utility method to create a node 
class newNode: 
  
    # Construct to create a new node 
    def __init__(self, key): 
        self.data = key 
        self.left = None
        self.right = None
  
# Iterative method to do level order 
# traversal line by line 
def printOddLevelEvenNodes(root):
    # Base Case 
    if (root == None):
        return
      
    # Create an empty queue for level 
    # order traversal 
    q =[] 
      
    # Enqueue root and initialize level as even
    q.append(root) 
    evenLevel = True
      
    while (1):
          
        # nodeCount (queue size) indicates 
        # number of nodes in the current level 
        nodeCount = len(q)
        if (nodeCount == 0):
            break
          
        # Mark 1st node as even positioned 
        evenNodePosition = True
          
        # Dequeue all the nodes of current level 
        # and Enqueue all the nodes of next level 
        while (nodeCount > 0):
            node = q[0]
            # Pronly even positioned 
            # nodes of even levels 
            if not evenLevel and evenNodePosition:
                print(node.data, end =" ")
            q.pop(0)
            if (node.left != None):
                q.append(node.left)
            if (node.right != None):
                q.append(node.right) 
            nodeCount-= 1
              
            # Switch the even position flag 
            evenNodePosition = not evenNodePosition 
          
        # Switch the even level flag 
        evenLevel = not evenLevel 
      
  
  
# Driver code 
if __name__ == '__main__': 
      
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.left = newNode(6)
    root.right.right = newNode(7)
    root.left.right.left = newNode(8)
    root.left.right.right = newNode(9)
    root.left.right.left.left = newNode(10)
    root.left.right.right.right = newNode(11)
  
    printOddLevelEvenNodes(root)


C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
{
class Node
{
    public int data;
    public Node left, right;
};
  
// Iterative method to do level order
// traversal line by line
static void printOddLevelEvenNodes(Node root)
{
    // Base Case
    if (root == null)
        return;
  
    // Create an empty queue for level
    // order traversal
    Queue q = new Queue();
  
    // Enqueue root and initialize level as even
    q.Enqueue(root);
    bool evenLevel = true;
  
    while (true) 
    {
  
        // nodeCount (queue size) indicates
        // number of nodes in the current level
        int nodeCount = q.Count;
        if (nodeCount == 0)
            break;
  
        // Mark 1st node as even positioned
        bool evenNodePosition = true;
  
        // Dequeue all the nodes of current level
        // and Enqueue all the nodes of next level
        while (nodeCount > 0) 
        {
            Node node = q.Peek();
  
            // Print only even positioned
            // nodes of even levels
            if (!evenLevel && evenNodePosition)
                Console.Write(node.data + " ");
            q.Dequeue();
            if (node.left != null)
                q.Enqueue(node.left);
            if (node.right != null)
                q.Enqueue(node.right);
            nodeCount--;
  
            // Switch the even position flag
            evenNodePosition = !evenNodePosition;
        }
  
        // Switch the even level flag
        evenLevel = !evenLevel;
    }
}
  
// Utility method to create a node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
  
// Driver code
public static void Main(String[] args) 
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.right.left = newNode(8);
    root.left.right.right = newNode(9);
    root.left.right.left.left = newNode(10);
    root.left.right.right.right = newNode(11);
  
    printOddLevelEvenNodes(root);
}
}
  
// This code is contributed by 29AjayKumar


输出:
2 8