+3 votes
109 views

Consider the following code

asked | 109 views
x is uninitialised..How to go about it..:)
x is no where used in the code. And the question asks for the sum of the internal nodes after the execution of that function. So i think ans is 22.
yes, x is not required in calculation . na?

@Habib

## 1 Answer

+5 votes
Best answer
It will change the value of all the nodes which are odd.

Means root->left  will become 0 from 1

and root->right will become 6 from 7

4

0 6

6 4 2

8 2 8

Now sum of internal nodes wil be : 4 + 0 + 6 + 6 + 4 + 2 =  22
answered by Loyal (2.8k points)
selected by
//check the code for the above , run it on ide ,
//and tested it yourself with different test cases.
#include <bits/stdc++.h>
using namespace std;
struct treenode{
int data;
treenode *left;
treenode *right;
};
treenode* create(int data){
treenode *temp = (treenode*)malloc(sizeof(treenode));
temp->data = data;
temp->left = temp->right  = NULL;
return temp;
}
int solve(treenode *root) { // will return 0 always
int x;
if(root == NULL) return 0;
else
if(root->data % 2 == 1){ // changing odd data items
root->data -= 1;
x++;
}
return solve(root->left) + solve(root->right);
}
void inorder(treenode *root) {
if(root == NULL) return;
inorder(root->left);
cout << root->data <<" ";
inorder(root->right);
}
int main(){
treenode *root;
root = create(4);
root->left = create(1);
root->right = create(7);
root->left->left = create(6);
root->left->right = create(4);
root->right->left = create(2);
root->left->left->right = create(8);
root->left->right->left = create(2);
root->right->left->left = create(8);
cout << solve(root) << endl;
inorder(root);
cout << endl;
return 0;
}

// output
0
6 8 0 2 4 4 8 2 6

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