# 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 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
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++ program for the above approach  ` ` `  `#include   ` `#include   ` `#include   ` `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 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 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# 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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