Delete Node in a BST

• traverse till the key matched,
• now have to take the decision whether to
  • make it null?
  • replace with left
  • replace with right
• can choose latter two, first one may lead to data lose.
• below implementation based upon replacing current one with the right
• but where to attach left?
• the leftmost of the right will be the smallest in so forming new Tree.
• therefore we should attach there.

 TreeNode* dfs(TreeNode* root, int key) {
        if (root == nullptr) {
            return root;
        }
        if (root->val == key) {
            if (root->right) {
                
                auto temp = root->right;
                while (temp->left != nullptr) temp = temp->left;
               temp->left = root->left;
                
                return root->right;
            }
            else {
                return root->left;
            }
            
        }
        
        root->left = dfs(root->left, key); 
        root->right = dfs(root->right, key); 
        return root;
    }
    
    TreeNode* deleteNode(TreeNode* root, int key) {
        root = dfs(root, key); 
        return root;
        
    }