Binary Tree Right Side View @LeetCode

PROBLEM :

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,
   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---
You should return [1, 3, 4].
   
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SIMPLE c++ IMPLEMENTATION :
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> vec ;
        if(root==NULL)
            return vec ;
           
        int height=heightTree(root) ;
        for(int i=0;i<height;i++){
            bool status=true ;
            rightView(root,vec,i,&status) ;
        }
        return vec ;
    }
   
    int heightTree(TreeNode* root){
        if(root==NULL)
            return 0 ;
        int l=heightTree(root->left) ;
        int r=heightTree(root->right) ;
        return l>r?l+1:r+1 ;
    }
   
    void rightView(TreeNode* root,vector<int> &vec,int level,bool *status){
        if(root==NULL||!(*status))
            return ;
           
        if(level==0){
            vec.push_back(root->val) ;
            (*status)=false ;
        }
        else{
            rightView(root->right,vec,level-1,&(*status)) ;
            rightView(root->left,vec,level-1,&(*status)) ;
        }
    }
};

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SIMPLE c++ IMPLEMENTATION : (Better Solution - Recursive)
--------------------------------------------------------------------------------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> vec ;
        RightView(root,vec,1) ;
        return vec ;
    }
   
    void RightView(TreeNode* root,vector<int> &vec,int level){
        if(root==NULL)
            return ;
           
        if(vec.size()<level)
            vec.push_back(root->val) ;
           
        RightView(root->right,vec,level+1) ;
        RightView(root->left,vec,level+1) ;
    }
};

--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION : (Better Solution - Iterative)
--------------------------------------------------------------------------------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> vec ;
        if(root==NULL)
            return vec ;
           
        queue<TreeNode*> que ;
        que.push(root) ;
        bool status ;
       
        while(!que.empty()){
            int size=que.size() ;
            status=true ;
           
            while(size--){
                TreeNode* curr=que.front() ;
                que.pop() ;
               
                if(status){
                    vec.push_back(curr->val) ;
                    status=false ;
                }
               
                if(curr->right)
                    que.push(curr->right) ;
                if(curr->left)
                    que.push(curr->left) ;
            }
        }
       
        return vec ;
    }
};

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