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.
Follow up: Can you solve it with time complexity O(height of tree)?
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: []
Approach :
Basically, the deletion can be divided into two stages:
10 10
/ \ delete(15) / \
7 15 ---------> 7 11
/ \ / / \
5 8 11 5 8
Solution:- Connect the prev child to next child
Nodes having 2 child / two subtree.
10 11
/ \ delete(10) / \
7 15 ---------> 7 15
/ \ / \ / \ \
5 8 11 18 5 8 18
Solution:-
1. Find the Inorder Predecessor(Max elemet of left subtree) or
2. Inorder Successor (Min element of the right subtree)
3. Replace with the Target node and delete the element.
Solution Iterative:
class Solution
{
public:
//finding min(every subtree of a BST is a BST
TreeNode *deleteNode(TreeNode *root, int key)
{
TreeNode *t = root;
TreeNode *q = NULL;
if (root == NULL)
{
return NULL;
}
while (t != NULL)
{
q = t;
if (t->val > key)
{
t = t->left;
}
else if (t->val < key)
{
t = t->right;
}
if (t != NULL && t->val == key)
{
//case 1: no child;
if (t->left == NULL && t->right == NULL)
{
if (q != t) // Checking if the node to be deleted is not root
{
if (q->left == t)
{
q->left = NULL;
}
else
{
q->right = NULL;
}
}
else
{
root = NULL;
}
}
//case 2: one child;
else if (t->left == NULL || t->right == NULL)
{
TreeNode *next = NULL; // Fidning the next node
if (t->left != NULL)
{
next = t->left;
}
else
{
next = t->right;
}
if (q != t) // Checking if the node to be deleted is not root
{
if (q->left == t)
{
q->left = next;
}
else
{
q->right = next;
}
}
else
{
root = next;
}
}
//case 3: two child
else
{
TreeNode *temp;
TreeNode *p = NULL;
temp = t->right;
// Finding the inorder succesor (i.e min element of right sub tree)
while (temp->left != NULL)
{
p = temp;
temp = temp->left;
}
if (p != NULL) // Checking if the node to be deleted is not root
{
p->left = temp->right;
}
else
{
t->right = temp->right;
}
t->val = temp->val;
}
break;
}
}
return root;
}
};
