### LeetCode Challenge #450. Delete Node in a BST

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:

1. Search for a node to remove.
2. 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.

```

Example 2:

```Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.
```

Example 3:

```Input: root = [], key = 0
Output: []
```

Constraints:

• The number of nodes in the tree is in the range `[0, 104]`.
• `-105 <= Node.val <= 105`
• Each node has a unique value.
• `root` is a valid binary search tree.
• `-105 <= key <= 105`
##### Video Solution
###### C++ Solution
```				```
class Solution {
public:
TreeNode* nodeDeleted(TreeNode* root, int key)
{
if(root==NULL)return root;
if(key<root->val)root->left=nodeDeleted(root->left,key);
else if(key>root->val)root->right=nodeDeleted(root->right,key);
else
{
if(!root->left and !root->right)return NULL;
else if(!root->left||!root->right) return root->left?root->left:root->right;

TreeNode* temp= root->right;
while(temp->left)temp=temp->left;
root->val=temp->val;
root->right=nodeDeleted(root->right,temp->val);
}
return root;
}
TreeNode* deleteNode(TreeNode* root, int key) {
return nodeDeleted(root,key);
}
};
```
```