1305.All_Elements_in_Two_Binary_Search_Trees.cpp

/*
    written by Pankaj Kumar.
    country:-INDIA
    Institute: National Institute of Technology, Uttarakhand
*/
typedef long long ll ;
typedef vector<int> vl;
typedef vector<vector<int>> vvl;
#define speed cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);
/*  Abbrevations  */
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
// loops
#define forin(arr,n) for(ll i=0;i<n;i++) cin>>arr[i];
// Some print
#define no cout<<"NO"<<endl;
#define yes cout<<"YES"<<endl;
// sort
#define all(V) (V).begin(),(V).end()
#define srt(V) sort(all(V))
#define srtGreat(V) sort(all(V),greater<ll>())
// some extra
#define printv(v) for(ll i=0;i<ll(v.size());i++){cout<<v[i]<<" ";} cout<<endl;
#define precision(x) cout<<fixed<<setprecision(x);
#define sz(V) ll(V.size())


/* ascii value
A=65,Z=90,a=97,z=122
*/

/* Some syntax 
//Syntax to create a min heap for priority queue
//priority_queue <int, vector<int>, greater<int>>pq;
*/
ll ppow(ll n, ll m, ll mod){
    if(m==0) return 1;
    ll tmp=ppow(n, m/2, mod);
    tmp=tmp*tmp%mod;
    return m%2 ? tmp*n%mod: tmp;
}
namespace mod_operations{
    ll modInv(ll n, ll mod){
        return ppow(n,mod-2, mod);
    }
    ll modAdd(ll n, ll m, ll mod){
        n=(n%mod+mod)%mod; m=(m%mod+mod)%mod;
        return (n+m)%mod;
    }
    ll modMul(ll n, ll m, ll mod){
        n=(n%mod+mod)%mod; m=(m%mod+mod)%mod;
        return n*m %mod;
    }
    ll modSub(ll n, ll m, ll mod){
        n=(n%mod+mod)%mod; m=(m%mod+mod)%mod;
        return modAdd(n,-m, mod);
    }
    ll modDiv(ll n, ll m, ll mod){
        return modMul(n, modInv(m, mod), mod);
    }
}
using namespace mod_operations;


/*  --------------------MAIN PROGRAM----------------------------*/
// to run ctrl+b
const ll INF=1e18;
const ll mod1=1e9+7;
const ll mod2=998244353;
// Techniques :
// divide into cases, brute force, pattern finding
// sort, greedy, binary search, two pointer
// transform into graph

// Experience :
// Cp is nothing but only observation and mathematics.


//Add main code here

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    map<int,int> m;
    void put_value(TreeNode* root){
        if(root==nullptr) return;
        m[root->val]++;
        put_value(root->left);
        put_value(root->right);
    }

    vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
        // for root1
        put_value(root1);
        // for root2
        put_value(root2);
        vector<int> ans;
        for(auto x:m){
            for(int i=1;i<=x.ss;i++){
                ans.pb(x.ff);
            }
        }
        
        return ans;
    }
};


/* -----------------END OF PROGRAM --------------------*/
/*
* stuff you should look before submission
* constraint and time limit
* int overflow
* special test case (n=0||n=1||n=2)
* don't get stuck on one approach if you get wrong answer
*/