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Complete below code of Preorder traversalPreorder(node* root){If(a-----------------){cout<<root->data;b--------------------------;preorder(root->right);}}                                                        a.NULL b.root->lefta.root==NULL b.Preorder(root->left);a.root!=NULL b.Preorder(root->left);a.root!=NULL b.root->left

Question

Complete below code of Preorder traversalPreorder(node* root){If(a-----------------){cout<<root->data;b--------------------------;preorder(root->right);}}                                                        a.NULL b.root->lefta.root==NULL b.Preorder(root->left);a.root!=NULL b.Preorder(root->left);a.root!=NULL b.root->left

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Solution

The correct completion of the code for Preorder traversal would be:

Preorder(node* root){
    If(root != NULL){
        cout<<root->data;
        Preorder(root->left);
        preorder(root->right);
    }
}

Here, the correct options for a and b are:

a. root != NULL b. Preorder(root->left)

This code first checks if the root is not NULL, then it prints the root data. After that, it recursively calls the Preorder function for the left subtree and then for the right subtree. This is the standard procedure for a Preorder traversal in a binary tree (root, left, right).

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Similar Questions

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ind a binary tree whose preorder and inorder traversals create thesame result

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