题目来源
删除二叉搜索树的一个节点。
题目确实不难，不过我写的有点乱，而且一开始没有考虑那么多东西，错了好多次才AC的，应该是有简洁一点的方法的。
我的代码如下：

/**
 * 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) {
        TreeNode *pre = NULL, *cur = root;
        while (cur != NULL) {
            if (cur->val == key) {
                if (cur->left == NULL) {
                    if (pre == NULL) {
                        delete cur;
                        return cur->right;
                    }
                    else {
                        if (pre->val > cur->val)
                            pre->left = cur->right;
                        else
                            pre->right = cur->right;
                        delete cur;
                        return root;
                    }
                }
                else {
                    if (cur->right == NULL) {
                        if (pre == NULL) {
                            delete cur;
                            return cur->left;
                        }
                        else {
                            if (pre->val > cur->val)
                                pre->left = cur->left;
                            else
                                pre->right = cur->left;
                            delete cur;
                            return root;
                        }
                    }
                    TreeNode *left_max = cur->left;
                    while (left_max->right)
                        left_max = left_max->right;
                    left_max->right = cur->right->left;
                    cur->right->left = cur->left;
                    if (pre == NULL) {
                        delete cur;
                        return cur->right;
                    }
                    else {
                        if (pre->val > cur->val)
                            pre->left = cur->right;
                        else
                            pre->right = cur->right;
                        delete cur;
                        return root;
                    }
                }
            }
            else if (cur->val > key) {
                pre = cur;
                cur = cur->left;
            }
            else {
                pre = cur;
                cur = cur->right;
            }
        }
        return root;
    }
};

看了看讨论区，可以用简洁一点的递归的方法来实现，代码如下：

/**
 * 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 == NULL)
            return root;
        if (key > root->val)
            root->right = deleteNode(root->right, key);
        else if (key < root->val)
            root->left = deleteNode(root->left, key);
        else {
            if (root->left == NULL)
                return root->right;
            else if (root->right == NULL)
                return root->left;
            root->val = findmin(root->right)->val;
            root->right = deleteNode(root->right, root->val);
        }
        return root;
    }
    
private:
    TreeNode *findmin(TreeNode *node)
    {
        while (node->left)
            node = node->left;
        return node;
    }
};