Find sum at level k

#1


getting wrong answer
I haven’t handled the case for 1 child
my doubt is, when node has 1 child how do we know that it is left or right child ?
and how to implement it ?

#2

Is your code failing all test cases ? .

#3

no
3 out of 4 failed, 1 passing

#4

I think ur not building the tree correctly. What about the NULL values after leaves … where have you appointed the NULL values.

#5

it gets initialized when new node is made, in node class I have initialized left and right child as NULL

#7

Done
Just made a little changes please check. Since the input is given in preorder form, then it means that if a node has 1 child it has got to be the left one.

#include<iostream>
using namespace std;

class node {
public:
    int data;
    node*left;
    node*right;
    node(int d):data(d),left(NULL),right(NULL) {}
};

node* build() {
    pair<int,int> p;
    cin>>p.first>>p.second;

    node* root=new node(p.first);
    if(p.second) {
        if(p.second>=1)
        root->left=build();
        if(p.second>=2)
        root->right=build();
    }
    return root;
}

int sum=0;
void getSumAtLevelK(node*root,int k) {    
    if(root==NULL)return ;
    if(k==0) {
        sum+=root->data;
        return;
    }
    getSumAtLevelK(root->left,k-1);
    getSumAtLevelK(root->right,k-1);
}

int main() {
    node*root=build();
    int k;
    cin>>k;

    getSumAtLevelK(root,k);
    cout<<sum<<"\n";

    return 0;
}
#8

Actually it doesnt matter if the node has only 1 child. You can make that either left or right. It doesn’t matter since we r dealing with levels only.

Please mark it resolved if ur doubt was cleared.

#9


getting run error

#10

I have added an extra condition please check the code more thoroughly :sweat_smile:

#11

@Madeshi-Chinmay-2159857244260221 got it, thank you

#12

If u feel ur doubt is resolved please rate ur experience.

#13

@Madeshi-Chinmay-2159857244260221 yes, already rated