delete-node-in-a-BST.cpp

/*
 
 PROBLEM STATEMENT
 
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.

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


Time Complexity : O(h) - h = height of the tree.
(Worst case Time Complexity : O(n) )

Code :

*/
TreeNode* deleteNode(TreeNode* root, int key) {
        if(root) 
            if(key < root->val) root->left = deleteNode(root->left, key);     //We frecursively call the function until we find the target node
            else if(key > root->val) root->right = deleteNode(root->right, key);       
            else{
                if(!root->left && !root->right) return NULL;          //No child condition
                if (!root->left || !root->right)
                    return root->left ? root->left : root->right;    //One child contion -> replace the node with it's child
					                                                //Two child condition   
                TreeNode* temp = root->left;                        //(or) TreeNode *temp = root->right;
                while(temp->right != NULL) temp = temp->right;     //      while(temp->left != NULL) temp = temp->left;
                root->val = temp->val;                            //       root->val = temp->val;
                root->left = deleteNode(root->left, temp->val);  //        root->right = deleteNode(root->right, temp);		
            }
        return root;
    }