text: Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.
  
  Basically, the deletion can be divided into two stages:
  
      Search for a node to remove.
      If the node is found, delete the node.
        
  Example 1:
  
  Input: root = [5,3,6,2,4,null,7], key = 3
  Output: [5,4,6,2,null,null,7]
  Explanation: Given key to delete is 3. So we find the node with value 3 and delete it.
  One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
  Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.
    
  /**
   * 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) {}
   * };
   */
    
  // recursive solution
  class Solution {
  public:
      TreeNode* deleteNode(TreeNode* root, int key) {
          if(root == nullptr){
              return root;
          }
          
          if(root->val == key){
              if(root->left == nullptr){
                  return root->right;
              }
              if(root->right == nullptr){
                  return root->left;
              }
              // the leftmost child node of right subtree
              TreeNode* cur = root->right;
              while(cur->left){
                  cur = cur->left;
              }
              // bind the left subtree to become the left child of cur
              cur->left = root->left;
              TreeNode* temp = root;
              // delete the node with the correct key value
              root = root->right;
              return root;
          }
          
          if(root->val > key){
              root->left = deleteNode(root->left, key);
          }
          if(root->val < key){
              root->right = deleteNode(root->right, key);
          }
          
          return root;
      }
  };
  