# Print all even nodes of Binary Search Tree

Given a binary search tree. The task is to print all even nodes of the binary search tree.

Examples:

Input :
5
/   \
3     7
/ \   / \
2   4 6   8
Output : 2 4 6 8

Input :
14
/   \
12    17
/ \   / \
8  13 16   19
Output : 8 12 14 16

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

Approach: Traverse the Binary Search tree and check if current node’s value is even. If yes then print it otherwise skip that node.

Below is the implementation of the above Approach:

 // C++ program to print all even node of BST #include using namespace std;    // create Tree struct Node {     int key;     struct Node *left, *right; };    // A utility function to create a new BST node Node* newNode(int item) {     Node* temp = new Node;     temp->key = item;     temp->left = temp->right = NULL;     return temp; }    // A utility function to do inorder traversal of BST void inorder(Node* root) {     if (root != NULL) {         inorder(root->left);         printf(root->key,end=" ");         inorder(root->right);     } }    /* A utility function to insert a new node    with given key in BST */ Node* insert(Node* node, int key) {     /* If the tree is empty, return a new node */     if (node == NULL)         return newNode(key);        /* Otherwise, recur down the tree */     if (key < node->key)         node->left = insert(node->left, key);     else         node->right = insert(node->right, key);        /* return the (unchanged) node pointer */     return node; }    // Function to print all even nodes void evenNode(Node* root) {     if (root != NULL) {         evenNode(root->left);         // if node is even then print it         if (root->key % 2 == 0)             printf("%d ", root->key);         evenNode(root->right);     } }    // Driver Code int main() {     /* Let us create following BST         5        / \       3     7      / \ / \      2 4 6 8 */     Node* root = NULL;     root = insert(root, 5);     root = insert(root, 3);     root = insert(root, 2);     root = insert(root, 4);     root = insert(root, 7);     root = insert(root, 6);     root = insert(root, 8);        evenNode(root);        return 0; }

 // Java program to print all even node of BST  class GfG {     // create Tree  static class Node {      int key;      Node left, right;  }    // A utility function to create a new BST node  static Node newNode(int item)  {      Node temp = new Node();      temp.key = item;      temp.left = null;     temp.right = null;      return temp;  }     // A utility function to do inorder traversal of BST  static void inorder(Node root)  {      if (root != null) {          inorder(root.left);          System.out.print(root.key + " ");          inorder(root.right);      }  }     /* A utility function to insert a new node  with given key in BST */ static Node insert(Node node, int key)  {      /* If the tree is empty, return a new node */     if (node == null)          return newNode(key);         /* Otherwise, recur down the tree */     if (key < node.key)          node.left = insert(node.left, key);      else         node.right = insert(node.right, key);         /* return the (unchanged) node pointer */     return node;  }     // Function to print all even nodes  static void evenNode(Node root)  {      if (root != null) {          evenNode(root.left);          // if node is even then print it          if (root.key % 2 == 0)              System.out.print(root.key + " ");          evenNode(root.right);      }  }     // Driver Code  public static void main(String[] args)  {      /* Let us create following BST      5      / \      3     7      / \ / \      2 4 6 8 */     Node root = null;      root = insert(root, 5);      root = insert(root, 3);      root = insert(root, 2);      root = insert(root, 4);      root = insert(root, 7);      root = insert(root, 6);      root = insert(root, 8);         evenNode(root);     } }

 # Python3 program to print all even node of BST     # create Tree  # to create a new BST node  class newNode:         # Construct to create a new node      def __init__(self, key):          self.key = key         self.left = None         self.right = None    # A utility function to do inorder  # traversal of BST  def inorder(root) :        if (root != None):          inorder(root.left)          printf("%d ", root.key)          inorder(root.right)         """ A utility function to insert a new  node with given key in BST """ def insert(node, key):         """ If the tree is empty,     return a new node """     if (node == None):          return newNode(key)         """ Otherwise, recur down the tree """     if (key < node.key):          node.left = insert(node.left, key)      else:         node.right = insert(node.right, key)         """ return the (unchanged)          node pointer """     return node     # Function to print all even nodes  def evenNode(root) :        if (root != None):          evenNode(root.left)                     # if node is even then print it          if (root.key % 2 == 0):             print(root.key, end = " ")          evenNode(root.right)     # Driver Code  if __name__ == '__main__':            """ Let us create following BST      5      / \      3 7      / \ / \      2 4 6 8 """     root = None     root = insert(root, 5)      root = insert(root, 3)      root = insert(root, 2)      root = insert(root, 4)      root = insert(root, 7)      root = insert(root, 6)      root = insert(root, 8)         evenNode(root)     # This code is contributed by # Shubham Singh(SHUBHAMSINGH10)

 // C# program to print all even node of BST  using System;    class GfG  {         // create Tree      class Node      {          public int key;          public Node left, right;      }         // A utility function to      // create a new BST node      static Node newNode(int item)      {          Node temp = new Node();          temp.key = item;          temp.left = null;          temp.right = null;          return temp;      }         // A utility function to do     // inorder traversal of BST      static void inorder(Node root)      {          if (root != null)          {              inorder(root.left);              Console.Write(root.key + " ");              inorder(root.right);          }      }         /* A utility function to insert a new node      with given key in BST */     static Node insert(Node node, int key)      {          /* If the tree is empty, return a new node */         if (node == null)              return newNode(key);             /* Otherwise, recur down the tree */         if (key < node.key)              node.left = insert(node.left, key);          else             node.right = insert(node.right, key);             /* return the (unchanged) node pointer */         return node;      }         // Function to print all even nodes      static void evenNode(Node root)      {          if (root != null)         {              evenNode(root.left);                             // if node is even then print it              if (root.key % 2 == 0)                  Console.Write(root.key + " ");              evenNode(root.right);          }      }         // Driver Code      public static void Main(String[] args)      {          /* Let us create following BST          5          / \          3 7          / \ / \          2 4 6 8 */         Node root = null;          root = insert(root, 5);          root = insert(root, 3);          root = insert(root, 2);          root = insert(root, 4);          root = insert(root, 7);          root = insert(root, 6);          root = insert(root, 8);             evenNode(root);      }  }     // This code has been contributed  // by PrinciRaj1992

Output:
2 4 6 8

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