检查二叉树中的两个节点是否是兄弟节点
给定一棵二叉树和两个节点，任务是检查节点是否是彼此的兄弟姐妹。
如果两个节点存在于同一级别，并且它们的父节点相同，则称它们为兄弟节点。
例子：
Input :
1
/ \
2 3
/ \ / \
4 5 6 7
First node is 4 and Second node is 6.
Output : No, they are not siblings.
Input :
1
/ \
5 6
/ / \
7 3 4
First node is 3 and Second node is 4
Output : Yes方法：仔细观察可以得出结论，二叉树中的任何节点最多可以有两个子节点。因此，由于两个兄弟姐妹的父节点必须相同，所以想法是简单地遍历树并为每个节点检查两个给定节点是否是它的子节点。如果树中的任何节点都为真，则打印 YES，否则打印 NO。
下面是上述方法的实现：
C++
// C++ program to check if two nodes are
// siblings
#include
using namespace std;
// Binary Tree Node
struct Node {
int data;
Node *left, *right;
};
// Utility function to create a new node
struct Node* newNode(int data)
{
struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Function to find out if two nodes are siblings
bool CheckIfNodesAreSiblings(Node* root, int data_one,
int data_two)
{
if (!root)
return false;
// Compare the two given nodes with
// the childrens of current node
if (root->left && root->right) {
int left = root->left->data;
int right = root->right->data;
if (left == data_one && right == data_two)
return true;
else if (left == data_two && right == data_one)
return true;
}
// Check for left subtree
if (root->left)
if(CheckIfNodesAreSiblings(root->left, data_one,
data_two))
return true;
// Check for right subtree
if (root->right)
if(CheckIfNodesAreSiblings(root->right, data_one,
data_two))
return true;
}
// Driver code
int main()
{
struct Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->right->left = newNode(5);
root->right->right = newNode(6);
root->left->left->right = newNode(7);
int data_one = 5;
int data_two = 6;
if (CheckIfNodesAreSiblings(root, data_one, data_two))
cout << "YES";
else
cout << "NO";
return 0;
} Java
// Java program to check if two nodes
// are siblings
import java.util.*;
class GFG{
// Binary Tree Node
static class Node
{
int data;
Node left, right;
};
// Utility function to create a
// new node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
static Node root = null;
// Function to find out if two nodes are siblings
static boolean CheckIfNodesAreSiblings(Node root,
int data_one,
int data_two)
{
if (root == null)
return false;
// Compare the two given nodes with
// the childrens of current node
if (root.left != null && root.right != null)
{
int left = root.left.data;
int right = root.right.data;
if (left == data_one &&
right == data_two)
return true;
else if (left == data_two &&
right == data_one)
return true;
}
// Check for left subtree
if (root.left != null)
CheckIfNodesAreSiblings(root.left,
data_one,
data_two);
// Check for right subtree
if (root.right != null)
CheckIfNodesAreSiblings(root.right,
data_one,
data_two);
return true;
}
// Driver code
public static void main(String[] args)
{
root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.right.left = newNode(5);
root.right.right = newNode(6);
root.left.left.right = newNode(7);
int data_one = 5;
int data_two = 6;
if (CheckIfNodesAreSiblings(root,
data_one,
data_two))
System.out.print("YES");
else
System.out.print("NO");
}
}
// This code is contributed by Rajput-JiPython3
# Python3 program to check if two
# nodes are siblings
# Binary Tree Node
class Node:
def __init__(self, key):
self.data = key
self.left = None
self.right = None
# Function to find out if two nodes are siblings
def CheckIfNodesAreSiblings(root, data_one,
data_two):
if (root == None):
return False
# Compare the two given nodes with
# the childrens of current node
ans = False
if (root.left != None and root.right != None):
left = root.left.data
right = root.right.data
if (left == data_one and right == data_two):
return True
elif (left == data_two and right == data_one):
return True
# Check for left subtree
if (root.left != None):
ans = ans or CheckIfNodesAreSiblings(root.left,
data_one,
data_two)
# Check for right subtree
if (root.right != None):
ans = ans or CheckIfNodesAreSiblings(root.right,
data_one,
data_two)
return ans
# Driver code
if __name__ == '__main__':
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.right.left = Node(5)
root.right.right = Node(6)
root.left.left.right = Node(7)
data_one = 5
data_two = 6
if (CheckIfNodesAreSiblings(root,
data_one,
data_two)):
print("YES")
else:
print("NO")
# This code is contributed by mohit kumar 29C#
// C# program to check if two nodes
// are siblings
using System;
// Binary Tree Node
public class Node
{
public int data;
public Node left, right;
// Utility function to create a
// new node
public Node(int item)
{
data = item;
left = right = null;
}
}
class GFG{
static Node root = null;
// Function to find out if two
// nodes are siblings
static bool CheckIfNodesAreSiblings(Node root,
int data_one,
int data_two)
{
if (root == null)
{
return false;
}
// Compare the two given nodes with
// the childrens of current node
if (root.left != null && root.right != null)
{
int left = root.left.data;
int right = root.right.data;
if (left == data_one && right == data_two)
{
return true;
}
else if (left == data_two && right == data_one)
{
return true;
}
}
// Check for left subtree
if (root.left != null)
{
CheckIfNodesAreSiblings(root.left, data_one,
data_two);
}
// Check for right subtree
if (root.right != null)
{
CheckIfNodesAreSiblings(root.right, data_one,
data_two);
}
return true;
}
// Driver code
static public void Main()
{
GFG.root = new Node(1);
GFG.root.left = new Node(2);
GFG.root.right = new Node(3);
GFG.root.left.left = new Node(4);
GFG.root.right.left = new Node(5);
GFG.root.right.right = new Node(6);
GFG.root.left.left.right = new Node(7);
int data_one = 5;
int data_two = 6;
if (CheckIfNodesAreSiblings(root, data_one,
data_two))
{
Console.WriteLine("YES");
}
else
{
Console.WriteLine("NO");
}
}
}
// This code is contributed by avanitrachhadiya2155Javascript
输出：
YES