Different Output while using and not using 'break' in switch case in c++

Question

I was solving the a question at HackerRank -
Question Link - https://www.hackerrank.com/challenges/maximum-element/problem

In one solution I used 'break' statement in switch-case, in another solution I didn't. Solution was wrong when I didn't use break statement. What is the reason behind this ?
Input -

10
1 97
2
1 20
2
1 26
1 20
2
3
1 91
3

With break Statement -

#include <bits/stdc++.h>
using namespace std;

int main() {
    int noOfTestCases;
    cin>>noOfTestCases;
    vector <int> st;
    for(int x=0; x<noOfTestCases; x++){
        int query;
        cin>>query;
        switch (query) {
            case 1:
                int number;
                cin>>number;
                if(st.empty()){
                    st.push_back(number);
                }
                else if(number > st[st.size()-1]){
                    st.push_back(number);
                }
                else{
                    st.push_back(st[st.size()-1]);
                }
                break;
            case 2:
                if(!st.empty()){
                    st.pop_back();
                }
                break;
            case 3:
                cout<<st[st.size()-1]<<endl;
        }
    }
}

//Output - 
//26
//91

Without break statement -

#include <bits/stdc++.h>
    using namespace std;

int main() {
    int noOfTestCases;
    cin>>noOfTestCases;
    vector <int> st;
    for(int x=0; x<noOfTestCases; x++){
        int query;
        cin>>query;
        switch (query) {
            case 1:
                int number;
                cin>>number;
                if(st.empty()){
                    st.push_back(number);
                }
                else if(number > st[st.size()-1]){
                    st.push_back(number);
                }
                else{
                    st.push_back(st[st.size()-1]);
                }
            case 2:
                if(!st.empty()){
                    st.pop_back();
                }
            case 3:
                cout<<st[st.size()-1]<<" "<<query<<endl;
        }
    }
}

//Output - 
//0
//0
//0
//0
//0
//0
//0
//0
//0
//0

Answer 1

So I had a misconception about switch case statement -

What I thought - I thought that in switch case if the expression is equal to any case D, then further cases - E, F, G, will not be executed, irrespective of the break statement.

Reality - If an expression is equal to any case D, then further cases - E, F, G will also be executed if we don't use break statement.

Thanks Jeffrey and Paul Sanders to clarify this problem.

Answer 2

Consider the following code

switch (x) {
  case 1:
    std::cout << "one\n";
  case 2:
    std::cout << "two\n";
    break;
  case 3:
    std::cout << "three\n";
}

If x is 1 it will print both one and two. It will then exit the switch block due to the break statement. Note that x will not be compared with 2 after printing "one", it will directly fall-through to printing "two".