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.
Note: Time complexity should be O(height of tree).
Example:
root = [5,3,6,2,4,null,7]
key = 3
5
/
3 6
/
2 4 7
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 following BST.
5
/
4 6
/
2 7
Another valid answer is [5,2,6,null,4,null,7].
5
/
2 6
4 7
Approach #1: C++.[recursive]
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* deleteNode(TreeNode* root, int key) {
if (root == nullptr) return nullptr;
if (root->val < key) root->right = deleteNode(root->right, key);
else if (root->val > key) root->left = deleteNode(root->left, key);
else {
if (root->left == nullptr) return root->right;
else if (root->right == nullptr) return root->left;
else {
TreeNode* minNode = findMinNode(root->right);
root->val = minNode->val;
root->right = deleteNode(root->right, root->val);
}
}
return root;
}
private:
TreeNode* findMinNode(TreeNode* node) {
if (node->left != nullptr) findMinNode(node->left);
return node;
}
};
Approach #2: Java.[recursive]
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null) return null;
if (root.val > key) root.left = deleteNode(root.left, key);
else if (root.val < key) root.right = deleteNode(root.right, key);
else {
if (root.left == null)
return root.right;
else if (root.right == null)
return root.left;
else {
TreeNode minNode = findMinNode(root.right);
root.val = minNode.val;
root.right = deleteNode(root.right, root.val);
}
}
return root;
}
private TreeNode findMinNode (TreeNode root) {
while (root.left != null) {
root = root.left;
}
return root;
}
}
Appraoch #3: Python.[Iterator]
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def deleteNode(self, root, key):
"""
:type root: TreeNode
:type key: int
:rtype: TreeNode
"""
cur = root
pre = None
while cur is not None and cur.val is not key:
pre = cur
if key < cur.val:
cur = cur.left
elif key > cur.val:
cur = cur.right
if pre is None:
return self.deleteRootNode(cur)
if pre.left == cur:
pre.left = self.deleteRootNode(cur)
else:
pre.right = self.deleteRootNode(cur)
return root
def deleteRootNode(self, node):
if node is None:
return None
if node.left is None:
return node.right
if node.right is None:
return node.left
nextNode = node.right
pre = None
while nextNode.left is not None:
pre = nextNode
nextNode = nextNode.left
nextNode.left = node.left
if node.right is not nextNode:
pre.left = nextNode.right
nextNode.right = node.right
return nextNode