# Count the number of visible nodes in Binary Tree

Given a Binary tree, the task is to find the number of visible nodes in the given binary tree. A node is a visible node if, in the path from the root to the node N, there is no node with greater value than N’s,

Examples:

```Input:
5
/  \
3     10
/  \   /
20   21 1

Output: 4
Explanation:
There are 4 visible nodes.
They are:
5: In the path 5 -> 3, 5 is the highest node value.
20: In the path 5 -> 3 -> 20, 20 is the highest node value.
21: In the path 5 -> 3 -> 21, 21 is the highest node value.
10: In the path 5 -> 10 -> 1, 10 is the highest node value.

Input:
-1
\
-2
\
-3
Output: 1
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to first traverse the tree. Since we need to see the maximum value in the given path, the pre-order traversal is used to traverse the given binary tree. While traversing the tree, we need to keep the track of the maximum value of the node that we have seen so far. If the current node is greater or equal to the max value, then increment the count of the visible node and update the max value with the current node value.

Below is the implementation of the above approach:

## Java

 `// Java implementation to count the ` `// number of visible nodes in ` `// the binary tree ` ` `  `// Class containing the left and right ` `// child of current node and the ` `// key value ` `class` `Node { ` `    ``int` `data; ` `    ``Node left, right; ` ` `  `    ``// Constructor of the class ` `    ``public` `Node(``int` `item) ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `public` `class` `GFG { ` `    ``Node root; ` ` `  `    ``// Variable to keep the track ` `    ``// of visible nodes ` `    ``static` `int` `count; ` ` `  `    ``// Function to perform the preorder traversal ` `    ``// for the given tree ` `    ``static` `void` `preOrder(Node node, ``int` `max) ` `    ``{ ` ` `  `        ``// Base case ` `        ``if` `(node == ``null``) { ` `            ``return``; ` `        ``} ` ` `  `        ``// If the current node value is greater ` `        ``// or equal to the max value, ` `        ``// then update count variable ` `        ``// and also update max variable ` `        ``if` `(node.data >= max) { ` `            ``count++; ` `            ``max = Math.max(node.data, max); ` `        ``} ` ` `  `        ``// Traverse to the left ` `        ``preOrder(node.left, max); ` ` `  `        ``// Traverse to the right ` `        ``preOrder(node.right, max); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``GFG tree = ``new` `GFG(); ` ` `  `        ``/* ` `                ``5 ` `               ``/  \ ` `             ``3     10 ` `            ``/  \   / ` `           ``20   21 1  ` `*/` ` `  `        ``tree.root = ``new` `Node(``5``); ` `        ``tree.root.left = ``new` `Node(``3``); ` `        ``tree.root.right = ``new` `Node(``10``); ` ` `  `        ``tree.root.left.left = ``new` `Node(``20``); ` `        ``tree.root.left.right = ``new` `Node(``21``); ` ` `  `        ``tree.root.right.left = ``new` `Node(``1``); ` ` `  `        ``count = ``0``; ` `        ``preOrder(tree.root, Integer.MIN_VALUE); ` ` `  `        ``System.out.println(count); ` `    ``} ` `} `

## C#

 `// C# implementation to count the ` `// number of visible nodes in ` `// the binary tree ` `using` `System; ` ` `  `// Class containing the left and right ` `// child of current node and the ` `// key value ` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node left, right; ` ` `  `    ``// Constructor of the class ` `    ``public` `Node(``int` `item) ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `class` `GFG{ ` `     `  `Node root; ` ` `  `// Variable to keep the track ` `// of visible nodes ` `static` `int` `count; ` ` `  `// Function to perform the preorder ` `// traversal for the given tree ` `static` `void` `preOrder(Node node, ``int` `max) ` `{ ` ` `  `    ``// Base case ` `    ``if` `(node == ``null``) ` `    ``{ ` `        ``return``; ` `    ``} ` ` `  `    ``// If the current node value is greater ` `    ``// or equal to the max value, ` `    ``// then update count variable ` `    ``// and also update max variable ` `    ``if` `(node.data >= max) ` `    ``{ ` `        ``count++; ` `        ``max = Math.Max(node.data, max); ` `    ``} ` ` `  `    ``// Traverse to the left ` `    ``preOrder(node.left, max); ` ` `  `    ``// Traverse to the right ` `    ``preOrder(node.right, max); ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main(String[] args) ` `{ ` `    ``GFG tree = ``new` `GFG(); ` ` `  `/* ` `         ``5 ` `        ``/ \ ` `      ``3       10 ` `     ``/ \   / ` `    ``20 21 1  ` `*/` ` `  `    ``tree.root = ``new` `Node(5); ` `    ``tree.root.left = ``new` `Node(3); ` `    ``tree.root.right = ``new` `Node(10); ` ` `  `    ``tree.root.left.left = ``new` `Node(20); ` `    ``tree.root.left.right = ``new` `Node(21); ` `    ``tree.root.right.left = ``new` `Node(1); ` ` `  `    ``count = 0; ` `    ``preOrder(tree.root, ``int``.MinValue); ` `    ``Console.WriteLine(count); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

Output:

```4
```

Complexity Analysis:
Time Complexity: O(N) where N is number of nodes in Binary tree.
Auxiliary Space: O(H) where H is the height of Binary tree.

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