1448.Count_Good_Nodes_in_Binary_Tree.cpp

/*
    written by Pankaj Kumar.
    country:-INDIA
    Institute: National Institute of Technology, Uttarakhand
*/
typedef long long ll ;
typedef unsigned long long ull;
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;
*/
/*  --------------------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:
    int good_nodes_count=0;
    void dfs(TreeNode* root,int maxo){
        if(root==nullptr){
            return ;
        }
        if(root->val>=maxo){
            good_nodes_count++;
        }
        maxo=max(maxo,root->val);
        dfs(root->left,maxo);
        dfs(root->right,maxo);
        return ;
    }

    int goodNodes(TreeNode* root) {
        int maxo=-1e5;
        dfs(root,maxo);
        return good_nodes_count;
    }
};



/* -----------------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
*/