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Check whether every node of binary tree has a value K on itself or its any immediate neighbours
  • Last Updated : 24 Aug, 2020

Given a binary tree and a value K, the task is to check if every node of the binary tree has either value of the node as K or at least one of its adjacent connected nodes has value K.
Examples: 
 

Input:
                     1
                   /   \
                  0     0
                /   \     \
               1     0     1
              /     /  \    \
             2     1    0    5
                  /    /
                 3    0
                     /
                    0
K = 0
Output: False
Explanation: 
We can observe that some leaf nodes
are having value other than 0 and
are not connected with any
adjacent node whose value is 0. 

Input:
                    0
                   / \
                  2   1
                       \
                        0
K = 0  
Output: True
Explanation: 
Since, all nodes of the tree
are either having value 0 or
connected with adjacent node
having value 0.

Approach: 

  1. The idea is to simply perform preorder traversal and pass reference of parent node recursively.
  2. While traversing for each node check the following conditions: 
    • if value of the node is K or,
    • if its parent node value is K or,
    • if any of its child node value is K.
  3. If any node of the tree doesn’t satisfy any of the given three conditions then, return False, else return True.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach 
  
#include <iostream> 
#include <unordered_map> 
#include <vector> 
using namespace std; 
  
// Defining tree node 
struct node { 
    int value; 
    struct node *right, *left; 
}; 
  
// Utility function to create 
// a new node 
struct node* newnode(int key) 
    struct node* temp = new node; 
    temp->value = key; 
    temp->right = NULL; 
    temp->left = NULL; 
  
    return temp; 
  
// Function to check binary 
// tree whether its every node 
// has value K or, it is 
// connected with node having 
// value K 
bool connectedK(node* root, 
                node* parent, 
                int K, 
                bool flag) 
    // checking node value 
    if (root->value != K) { 
  
        // checking the left 
        // child value 
        if (root->left == NULL 
            || root->left->value != K) { 
  
            // checking the rigth 
            // child value 
            if (root->right == NULL 
                || root->right->value != K) { 
  
                // checking the parent value 
                if (parent == NULL 
                    || parent->value != K) { 
                    flag = false
                    return flag; 
                
            
        
    
  
    // Traversing to the left subtree 
    if (root->left != NULL) { 
        if (flag == true) { 
            flag 
                = connectedK(root->left, 
                            root, K, flag); 
        
    
  
    // Traversing to the right subtree 
    if (root->right != NULL) { 
        if (flag == true) { 
            flag 
                = connectedK(root->right, 
                            root, K, flag); 
        
    
    return flag; 
  
// Driver code 
int main() 
  
    // Input Binary tree 
    struct node* root = newnode(0); 
    root->right = newnode(1); 
    root->right->right = newnode(0); 
    root->left = newnode(0); 
  
    int K = 0; 
  
    // Function call to check 
    // binary tree 
    bool result 
        = connectedK(root, NULL, 
                    K, true); 
  
    if (result == false
        cout << "False\n"
    else
        cout << "True\n"
  
    return 0; 

Java




// Java program for the above approach 
import java.util.*;
class GFG{ 
  
// Defining tree node 
static class node
    int value; 
    node right, left; 
}; 
  
// Utility function to create 
// a new node 
static node newnode(int key) 
    node temp = new node(); 
    temp.value = key; 
    temp.right = null
    temp.left = null
  
    return temp; 
  
// Function to check binary 
// tree whether its every node 
// has value K or, it is 
// connected with node having 
// value K 
static boolean connectedK(node root, 
                          node parent, 
                          int K, 
                          boolean flag) 
    // Checking node value 
    if (root.value != K) 
    
  
        // Checking the left 
        // child value 
        if (root.left == null ||
            root.left.value != K)
        
  
            // Checking the rigth 
            // child value 
            if (root.right == null ||
                root.right.value != K) 
            
  
                // Checking the parent value 
                if (parent == null || 
                    parent.value != K)
                
                    flag = false
                    return flag; 
                
            
        
    
  
    // Traversing to the left subtree 
    if (root.left != null
    
        if (flag == true)
        
            flag = connectedK(root.left, 
                              root, K, flag); 
        
    
  
    // Traversing to the right subtree 
    if (root.right != null
    
        if (flag == true
        
            flag = connectedK(root.right, 
                              root, K, flag); 
        
    
    return flag; 
  
// Driver code 
public static void main(String[] args) 
  
    // Input Binary tree 
    node root = newnode(0); 
    root.right = newnode(1); 
    root.right.right = newnode(0); 
    root.left = newnode(0); 
  
    int K = 0
  
    // Function call to check 
    // binary tree 
    boolean result = connectedK(root, null
                                   K, true); 
  
    if (result == false
        System.out.print("False\n"); 
    else
        System.out.print("True\n"); 
  
// This code is contributed by Rohit_ranjan

Python3




# Python3 program for the above approach
  
# Defining tree node
class Node:
    def __init__(self,key):
          
        self.value = key
        self.left = None
        self.right = None
          
# Function to check binary 
# tree whether its every node 
# has value K or, it is 
# connected with node having 
# value K 
def connectedK(root, parent, K, flag):
      
    # Checking node value
    if root.value != K:
          
        # Checking the left 
        # child value 
        if (root.left == None or 
            root.left.value != K):
              
            # Checking the right 
            # child value 
            if (root.right == None or 
                root.right.value != K):
                  
                # Checking the parent value 
                if (parent == None or 
                    parent.value != K):
                    flag = False
                    return flag
                  
    # Traversing to the left subtree
    if root.left != None:
        if flag == True:
            flag = connectedK(root.left, 
                              root, K, flag)
      
    # Traversing to the right subtree
    if root.right != None:
        if flag == True:
            flag = connectedK(root.right, 
                              root, K, flag)
              
    return flag
  
# Driver code
root = Node(0)
root.right = Node(1)
root.right.right = Node(0)
root.left = Node(0)
  
K = 0
  
# Function call to check 
# binary tree 
result = connectedK(root, None, K, True)
  
if result == False:
    print("False")
else:
    print("True")
  
# This code is contributed by Stuti Pathak

C#




// C# program for the above approach 
using System;
  
class GFG{ 
  
// Defining tree node 
class node
    public int value; 
    public node right, left; 
}; 
  
// Utility function to create 
// a new node 
static node newnode(int key) 
    node temp = new node(); 
    temp.value = key; 
    temp.right = null
    temp.left = null
  
    return temp; 
  
// Function to check binary 
// tree whether its every node 
// has value K or, it is 
// connected with node having 
// value K 
static bool connectedK(node root, 
                       node parent, 
                       int K, 
                       bool flag) 
      
    // Checking node value 
    if (root.value != K) 
    
  
        // checking the left 
        // child value 
        if (root.left == null ||
            root.left.value != K)
        
  
            // Checking the rigth 
            // child value 
            if (root.right == null ||
                root.right.value != K) 
            
  
                // Checking the parent value 
                if (parent == null || 
                    parent.value != K)
                
                    flag = false
                    return flag; 
                
            
        
    
  
    // Traversing to the left subtree 
    if (root.left != null
    
        if (flag == true)
        
            flag = connectedK(root.left, 
                              root, K, flag); 
        
    
  
    // Traversing to the right subtree 
    if (root.right != null
    
        if (flag == true
        
            flag = connectedK(root.right, 
                              root, K, flag); 
        
    
    return flag; 
  
// Driver code 
public static void Main(String[] args) 
  
    // Input Binary tree 
    node root = newnode(0); 
    root.right = newnode(1); 
    root.right.right = newnode(0); 
    root.left = newnode(0); 
  
    int K = 0; 
  
    // Function call to check 
    // binary tree 
    bool result = connectedK(root, null
                                K, true); 
  
    if (result == false
        Console.Write("False\n"); 
    else
        Console.Write("True\n"); 
  
// This code is contributed by Princi Singh

Output: 

True

Time complexity: O(N), N is the number of nodes of the tree. 
Auxiliary Space: O(1)
 




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