LC 450. Delete Node in a BST
2024-09-24 08:11:44
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
class Solution {
public:
TreeNode* deleteNode(TreeNode* root, int key) {
if(!root) return root;
TreeNode* ret;
if(root->val == key) {
TreeNode* rnode_lmost = getlm_or_rm_node(root->right, true);
if(rnode_lmost) {
rnode_lmost->left = root->left;
ret = root->right;
}else ret = root->left;
}else {
if(key < root->val) root->left = deleteNode(root->left, key);
else root->right = deleteNode(root->right, key);
ret = root;
}
return ret;
}
TreeNode* getlm_or_rm_node(TreeNode* root, bool left){
if(!root) return root;
if(left) {
while(root->left) root = root->left;
}else {
while(root->right) root = root->right;
}
return root;
}
};
class Solution {
public:
TreeNode *deleteNode(TreeNode *root, int key) {
TreeNode **cur = &root;
while (*cur && (*cur)->val != key)
cur = (key > (*cur)->val) ? &(*cur)->right : &(*cur)->left;
if (*cur) {
if (!(*cur)->right) *cur = (*cur)->left;
else {
TreeNode **successor = &(*cur)->right;
while ((*successor)->left) successor = &(*successor)->left;
swap((*cur)->val, (*successor)->val);
*successor = (*successor)->right ? (*successor)->right : nullptr;
}
}
return root;
}
};
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