LeetCode 450 Delete Node in a BST（删除BST节点）

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

题目大意：删除BST上的节点。

代码如下：

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     struct TreeNode *left;
*     struct TreeNode *right;
* };
*/

struct TreeNode* findMax(struct TreeNode* root)
{
if(root == NULL) return NULL;
else if(root->right == NULL) return root;
else return findMax(root->right);
}

struct TreeNode* deleteNode(struct TreeNode* root, int key) {
if(root == NULL) return NULL;
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 && root->right){
struct TreeNode* tmp = findMax(root->left);
root->val = tmp->val;
root->left = deleteNode(root->left, tmp->val);
}else{
struct TreeNode* tmp = root;
if(root->left == NULL) root = root->right;
else if(root->right == NULL) root = root->left;
free(tmp);
}
return root;
}