在二叉树中打印给定节点的表兄弟

给定一棵二叉树和一个节点，打印给定节点的所有表兄弟。请注意，不应打印兄弟姐妹。
例子：

Input : root of below tree 
             1
           /   \
          2     3
        /   \  /  \
       4    5  6   7
       and pointer to a node say 5.

Output : 6, 7

使用此处讨论的方法首先找到给定节点级别的想法。找到级别后，我们可以使用此处讨论的方法打印给定级别的所有节点。唯一需要注意的是，兄弟姐妹不应该被打印。为了处理这个问题，我们将打印函数更改为首先检查兄弟节点并仅在它不是兄弟节点时才打印节点。
下面是上述想法的实现。

C++

// C++ program to print cousins of a node
#include 
using namespace std;
 
// A Binary Tree Node
struct Node
{
    int data;
    Node *left, *right;
};
 
// A utility function to create a new
// Binary Tree Node
Node *newNode(int item)
{
    Node *temp = new Node;
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}
 
/* It returns level of the node if it is
present in tree, otherwise returns 0.*/
int getLevel(Node *root, Node *node, int level)
{
     
    // base cases
    if (root == NULL)
        return 0;
    if (root == node)
        return level;
 
    // If node is present in left subtree
    int downlevel = getLevel(root->left,
                             node, level + 1);
    if (downlevel != 0)
        return downlevel;
 
    // If node is not present in left subtree
    return getLevel(root->right, node, level + 1);
}
 
/* Print nodes at a given level such that
sibling of node is not printed if it exists */
void printGivenLevel(Node* root, Node *node, int level)
{
    // Base cases
    if (root == NULL || level < 2)
        return;
 
    // If current node is parent of a node
    // with given level
    if (level == 2)
    {
        if (root->left == node || root->right == node)
            return;
        if (root->left)
            cout << root->left->data << " ";
        if (root->right)
            cout << root->right->data;
    }
 
    // Recur for left and right subtrees
    else if (level > 2)
    {
        printGivenLevel(root->left, node, level - 1);
        printGivenLevel(root->right, node, level - 1);
    }
}
 
// This function prints cousins of a given node
void printCousins(Node *root, Node *node)
{
    // Get level of given node
    int level = getLevel(root, node, 1);
 
    // Print nodes of given level.
    printGivenLevel(root, node, level);
}
 
// Driver Code
int main()
{
    Node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->left->right->right = newNode(15);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
 
    printCousins(root, root->left->right);
 
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

// C program to print cousins of a node
#include 
#include 
 
// A Binary Tree Node
struct Node
{
    int data;
    Node *left, *right;
};
 
// A utility function to create a new Binary
// Tree Node
Node *newNode(int item)
{
    Node *temp =  new Node;
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}
 
/* It returns level of the node if it is present
   in tree, otherwise returns 0.*/
int getLevel(Node *root, Node *node, int level)
{
    // base cases
    if (root == NULL)
        return 0;
    if (root == node)
        return level;
 
    // If node is present in left subtree
    int downlevel = getLevel(root->left, node, level+1);
    if (downlevel != 0)
        return downlevel;
 
    // If node is not present in left subtree
    return getLevel(root->right, node, level+1);
}
 
/* Print nodes at a given level such that sibling of
   node is not printed if it exists  */
void printGivenLevel(Node* root, Node *node, int level)
{
    // Base cases
    if (root == NULL || level < 2)
        return;
 
    // If current node is parent of a node with
    // given level
    if (level == 2)
    {
        if (root->left == node || root->right == node)
            return;
        if (root->left)
           printf("%d ", root->left->data);
        if (root->right)
           printf("%d ", root->right->data);
    }
 
    // Recur for left and right subtrees
    else if (level > 2)
    {
        printGivenLevel(root->left, node, level-1);
        printGivenLevel(root->right, node, level-1);
    }
}
 
// This function prints cousins of a given node
void printCousins(Node *root, Node *node)
{
    // Get level of given node
    int level = getLevel(root, node, 1);
 
    // Print nodes of given level.
    printGivenLevel(root, node, level);
}
 
// Driver Program to test above functions
int main()
{
    Node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->left->right->right = newNode(15);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
 
    printCousins(root, root->left->right);
 
    return 0;
}

Java

// Java program to print cousins of a node
class GfG {
 
// A Binary Tree Node
static class Node
{
    int data;
    Node left, right;
}
 
// A utility function to create a new Binary
// Tree Node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.left = null;
    temp.right = null;
    return temp;
}
 
/* It returns level of the node if it is present
in tree, otherwise returns 0.*/
static int getLevel(Node root, Node node, int level)
{
    // base cases
    if (root == null)
        return 0;
    if (root == node)
        return level;
 
    // If node is present in left subtree
    int downlevel = getLevel(root.left, node, level+1);
    if (downlevel != 0)
        return downlevel;
 
    // If node is not present in left subtree
    return getLevel(root.right, node, level+1);
}
 
/* Print nodes at a given level such that sibling of
node is not printed if it exists */
static void printGivenLevel(Node root, Node node, int level)
{
    // Base cases
    if (root == null || level < 2)
        return;
 
    // If current node is parent of a node with
    // given level
    if (level == 2)
    {
        if (root.left == node || root.right == node)
            return;
        if (root.left != null)
        System.out.print(root.left.data + " ");
        if (root.right != null)
        System.out.print(root.right.data + " ");
    }
 
    // Recur for left and right subtrees
    else if (level > 2)
    {
        printGivenLevel(root.left, node, level-1);
        printGivenLevel(root.right, node, level-1);
    }
}
 
// This function prints cousins of a given node
static void printCousins(Node root, Node node)
{
    // Get level of given node
    int level = getLevel(root, node, 1);
 
    // Print nodes of given level.
    printGivenLevel(root, node, level);
}
 
// Driver Program to test above functions
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.left.right.right = newNode(15);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
 
    printCousins(root, root.left.right);
}
}

Python3

# Python3 program to print cousins of a node
 
# A utility function to create a new
# Binary Tree Node
class newNode:
    def __init__(self, item):
        self.data = item
        self.left = self.right = None
 
# It returns level of the node if it is
# present in tree, otherwise returns 0.
def getLevel(root, node, level):
     
    # base cases
    if (root == None):
        return 0
    if (root == node):
        return level
 
    # If node is present in left subtree
    downlevel = getLevel(root.left, node,
                               level + 1)
    if (downlevel != 0):
        return downlevel
 
    # If node is not present in left subtree
    return getLevel(root.right, node, level + 1)
 
# Print nodes at a given level such that
# sibling of node is not printed if
# it exists
def printGivenLevel(root, node, level):
     
    # Base cases
    if (root == None or level < 2):
        return
 
    # If current node is parent of a
    # node with given level
    if (level == 2):
        if (root.left == node or
            root.right == node):
            return
        if (root.left):
            print(root.left.data, end = " ")
        if (root.right):
            print(root.right.data, end = " ")
 
    # Recur for left and right subtrees
    else if (level > 2):
        printGivenLevel(root.left, node, level - 1)
        printGivenLevel(root.right, node, level - 1)
 
# This function prints cousins of a given node
def printCousins(root, node):
     
    # Get level of given node
    level = getLevel(root, node, 1)
 
    # Print nodes of given level.
    printGivenLevel(root, node, level)
 
# 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.left.right.right = newNode(15)
    root.right.left = newNode(6)
    root.right.right = newNode(7)
    root.right.left.right = newNode(8)
 
    printCousins(root, root.left.right)
     
# This code is contributed by PranchalK

C#

// C# program to print cousins of a node
using System;
 
public class GfG
{
 
// A Binary Tree Node
class Node
{
    public int data;
    public Node left, right;
}
 
// A utility function to create 
// a new Binary Tree Node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.left = null;
    temp.right = null;
    return temp;
}
 
/* It returns level of the node
if it is present in tree,
 otherwise returns 0.*/
static int getLevel(Node root,
            Node node, int level)
{
    // base cases
    if (root == null)
        return 0;
    if (root == node)
        return level;
 
    // If node is present in left subtree
    int downlevel = getLevel(root.left, node, level + 1);
    if (downlevel != 0)
        return downlevel;
 
    // If node is not present in left subtree
    return getLevel(root.right, node, level + 1);
}
 
/* Print nodes at a given level
such that sibling of node is
 not printed if it exists */
static void printGivenLevel(Node root,
                    Node node, int level)
{
    // Base cases
    if (root == null || level < 2)
        return;
 
    // If current node is parent of a node with
    // given level
    if (level == 2)
    {
        if (root.left == node || root.right == node)
            return;
        if (root.left != null)
            Console.Write(root.left.data + " ");
        if (root.right != null)
            Console.Write(root.right.data + " ");
    }
 
    // Recur for left and right subtrees
    else if (level > 2)
    {
        printGivenLevel(root.left, node, level - 1);
        printGivenLevel(root.right, node, level - 1);
    }
}
 
// This function prints cousins of a given node
static void printCousins(Node root, Node node)
{
    // Get level of given node
    int level = getLevel(root, node, 1);
 
    // Print nodes of given level.
    printGivenLevel(root, node, level);
}
 
// 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.left.right.right = newNode(15);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
 
    printCousins(root, root.left.right);
}
}
 
// This code is contributed Rajput-Ji

Javascript

  
// JavaScript program to print cousins of a node
 
// A Binary Tree Node
class Node
{
  constructor()
  {
    this.data=0;
    this.left= null;
    this.right = null;
  }
}
 
// A utility function to create 
// a new Binary Tree Node
function newNode(item)
{
    var temp = new Node();
    temp.data = item;
    temp.left = null;
    temp.right = null;
    return temp;
}
 
/* It returns level of the node
if it is present in tree,
 otherwise returns 0.*/
function getLevel(root, node, level)
{
    // base cases
    if (root == null)
        return 0;
    if (root == node)
        return level;
 
    // If node is present in left subtree
    var downlevel = getLevel(root.left, node, level + 1);
    if (downlevel != 0)
        return downlevel;
 
    // If node is not present in left subtree
    return getLevel(root.right, node, level + 1);
}
 
/* Print nodes at a given level
such that sibling of node is
 not printed if it exists */
function printGivenLevel(root, node, level)
{
    // Base cases
    if (root == null || level < 2)
        return;
 
    // If current node is parent of a node with
    // given level
    if (level == 2)
    {
        if (root.left == node || root.right == node)
            return;
        if (root.left != null)
            document.write(root.left.data + " ");
        if (root.right != null)
            document.write(root.right.data + " ");
    }
 
    // Recur for left and right subtrees
    else if (level > 2)
    {
        printGivenLevel(root.left, node, level - 1);
        printGivenLevel(root.right, node, level - 1);
    }
}
 
// This function prints cousins of a given node
function printCousins(root, node)
{
    // Get level of given node
    var level = getLevel(root, node, 1);
 
    // Print nodes of given level.
    printGivenLevel(root, node, level);
}
 
// Driver code
var root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.left.right.right = newNode(15);
root.right.left = newNode(6);
root.right.right = newNode(7);
root.right.left.right = newNode(8);
printCousins(root, root.left.right);
 

输出 ：

6 7