Check whether every node of binary tree has a value K on itself or its any immediate neighbours

Given a binary tree and a value K, the task is to check if every node of the binary tree has either the value of the node as K or at least one of its adjacent connected nodes has the 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 the reference of parent node recursively.
2. While traversing for each node check the following conditions: 
   • if the 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 right
            // 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 right
            // 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 right
            // 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

Javascript

<script>
    // Javascript program for the above approach

    // Defining tree node
    class node
    {
        constructor(key) {
           this.value = key;
           this.left = this.right = null;
        }
    }

    // Utility function to create
    // a new node
    function newnode(key)
    {
        let temp = new node(key);
        return temp;
    }

    // Function to check binary
    // tree whether its every node
    // has value K or, it is
    // connected with node having
    // value K
    function connectedK(root, parent, K, flag)
    {
        // Checking node value
        if (root.value != K)
        {

            // Checking the left
            // child value
            if (root.left == null ||
                root.left.value != K)
            {

                // Checking the right
                // 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;
    }

    // Input Binary tree
    let root = newnode(0);
    root.right = newnode(1);
    root.right.right = newnode(0);
    root.left = newnode(0);

    let K = 0;

    // Function call to check
    // binary tree
    let result = connectedK(root, null, K, true);

    if (result == false)
        document.write("False");
    else
        document.write("True");

</script>

Output:

True

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