Delete Node in a BST
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
Code
Python Code
class Solution(object):
def deleteNode(self, root, key):
if not root: return None
if root.val == key:
if not root.right: return root.left
if not root.left: return root.right
if root.left and root.right:
temp = root.right
while temp.left: temp = temp.left
root.val = temp.val
root.right = self.deleteNode(root.right, root.val)
elif root.val > key:
root.left = self.deleteNode(root.left, key)
else:
root.right = self.deleteNode(root.right, key)
return root
C++ Code
class Solution {
public:
TreeNode* deleteNode(TreeNode* root, int key) {
if(root)
if(key < root->val) root->left = deleteNode(root->left, key);
else if(key > root->val) root->right = deleteNode(root->right, key);
else{
if(!root->left && !root->right) return NULL;
if (!root->left || !root->right)
return root->left ? root->left : root->right;
TreeNode* temp = root->left;
while(temp->right ! = NULL) temp = temp->right;
root->val = temp->val;
root->left = deleteNode(root->left, temp->val);
}
return root;
}
};