打印有孙子的二叉树的节点

给定一个二叉树，任务是打印有孙子节点的节点。

例子：

Input: 
 

Output: 20 8 
Explanation: 
20 and 8 are the grandparents of 4, 12 and 10, 14.

Input: 
 

Output: 1 
Explanation: 
1 is the grandparent of 4, 5. 

方法：这个想法使用递归。以下是步骤：

1. 在每个节点遍历给定的树。
2. 检查每个节点是否有孙节点。
3. 对于任何树节点(比如temp )，如果存在以下节点之一，则当前节点是祖父节点：
   • 温度->左->左。
   • 温度->左->右。
   • 温度-> 右-> 左。
   • 温度-> 右-> 右。
4. 如果任何节点temp存在上述任何一项，则节点temp是祖父节点。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// A Binary Tree Node
struct node {
    struct node *left, *right;
    int key;
};
 
// Function to create new tree node
node* newNode(int key)
{
    node* temp = new node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Function to print the nodes of
// the Binary Tree having a grandchild
void cal(struct node* root)
{
    // Base case to check
    // if the tree exists
 
    if (root == NULL)
        return;
 
    else {
 
        // Check if there is a left and
        // right child of the curr node
        if (root->left != NULL
            && root->right != NULL) {
 
            // Check for grandchildren
            if (root->left->left != NULL
                || root->left->right != NULL
                || root->right->left != NULL
                || root->right->right != NULL) {
 
                // Print the node's key
                cout << root->key << " ";
            }
        }
 
        // Check if the left child
        // of node is not null
        else if (root->left != NULL) {
 
            // Check for grandchildren
            if (root->left->left != NULL
                || root->left->right != NULL) {
                cout << root->key << " ";
            }
        }
 
        // Check if the right child
        // of node is not null
        else if (root->right != NULL) {
 
            // Check for grandchildren
            if (root->right->left != NULL
                || root->right->right != NULL) {
                cout << root->key << " ";
            }
        }
 
        // Recursive call on left and
        // right subtree
        cal(root->left);
        cal(root->right);
    }
}
 
// Driver Code
int main()
{
    // Given Tree
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    // Function Call
    cal(root);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// A Binary Tree Node
static class node
{
    node left, right;
    int key;
};
 
// Function to create new tree node
static node newNode(int key)
{
    node temp = new node();
    temp.key = key;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to print the nodes of
// the Binary Tree having a grandchild
static void cal(node root)
{
     
    // Base case to check
    // if the tree exists
    if (root == null)
        return;
 
    else
    {
         
        // Check if there is a left and
        // right child of the curr node
        if (root.left != null &&
           root.right != null)
        {
             
            // Check for grandchildren
            if (root.left.left != null ||
               root.left.right != null ||
               root.right.left != null ||
              root.right.right != null)
            {
                 
                // Print the node's key
                System.out.print(root.key + " ");
            }
        }
 
        // Check if the left child
        // of node is not null
        else if (root.left != null)
        {
             
            // Check for grandchildren
            if (root.left.left != null ||
               root.left.right != null)
            {
                System.out.print(root.key + " ");
            }
        }
 
        // Check if the right child
        // of node is not null
        else if (root.right != null)
        {
             
            // Check for grandchildren
            if (root.right.left != null ||
               root.right.right != null)
            {
                System.out.print(root.key + " ");
            }
        }
 
        // Recursive call on left and
        // right subtree
        cal(root.left);
        cal(root.right);
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Tree
    node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
 
    // Function call
    cal(root);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the
# above approach
 
# A Binary Tree Node
class newNode:
   
    def __init__(self, key):
       
        self.key = key
        self.left = None
        self.right = None
 
# Function to print the nodes
# of the Binary Tree having a
# grandchild
def cal(root):
   
    # Base case to check
    # if the tree exists
    if (root == None):
        return
 
    else:
        # Check if there is a left
        # and right child of the
        # curr node
        if (root.left != None and
            root.right != None):
           
            # Check for grandchildren
            if (root.left.left != None or
                root.left.right != None or
                root.right.left != None or
                root.right.right != None):
               
                # Print the node's key
                print(root.key, end = " ")
 
        # Check if the left child
        # of node is not None
        elif (root.left != None):
           
            # Check for grandchildren
            if (root.left.left != None or
                root.left.right != None):
                print(root.key, end = " ")
 
        # Check if the right child
        # of node is not None
        elif(root.right != None):
           
            # Check for grandchildren
            if (root.right.left != None or
                root.right.right != None):
                print(root.key, end = " ")
 
        # Recursive call on left and
        # right subtree
        cal(root.left)
        cal(root.right)
 
# Driver Code
if __name__ == '__main__':
   
    # Given Tree
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
 
    # Function Call
    cal(root)
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the
// above approach
using System;
class GFG{
 
// A Binary Tree Node
public class node
{
  public node left, right;
  public int key;
};
 
// Function to create new
// tree node
static node newNode(int key)
{
  node temp = new node();
  temp.key = key;
  temp.left = temp.right = null;
  return temp;
}
 
// Function to print the
// nodes of the Binary Tree
// having a grandchild
static void cal(node root)
{
  // Base case to check
  // if the tree exists
  if (root == null)
    return;
   
  else
  {
    // Check if there is a left and
    // right child of the curr node
    if (root.left != null &&
        root.right != null)
    {
      // Check for grandchildren
      if (root.left.left != null ||
          root.left.right != null ||
          root.right.left != null ||
          root.right.right != null)
      {
        // Print the node's key
        Console.Write(root.key + " ");
      }
    }
 
    // Check if the left child
    // of node is not null
    else if (root.left != null)
    {
      // Check for grandchildren
      if (root.left.left != null ||
          root.left.right != null)
      {
        Console.Write(root.key + " ");
      }
    }
 
    // Check if the right child
    // of node is not null
    else if (root.right != null)
    {
      // Check for grandchildren
      if (root.right.left != null ||
          root.right.right != null)
      {
        Console.Write(root.key + " ");
      }
    }
 
    // Recursive call on left and
    // right subtree
    cal(root.left);
    cal(root.right);
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given Tree
  node root = newNode(1);
  root.left = newNode(2);
  root.right = newNode(3);
  root.left.left = newNode(4);
  root.left.right = newNode(5);
 
  // Function call
  cal(root);
}
}
 
// This code is contributed by Princi Singh

输出：

1

时间复杂度： O(N) ，其中 N 是节点数。
辅助空间： O(1)

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