450. 删除二叉搜索树中的节点

题目链接：https://leetcode.cn/problems/delete-node-in-a-bst/

这个题目自己没做出来，看的官方题解，最后一步successor找到之后再删除最后返回新的root这段没怎么看明白。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) {
            return nullptr;
        }

        if (root->val > key) {
            root->left = deleteNode(root->left, key);
            return root;
        }

        if (root->val < key) {
            root->right = deleteNode(root->right, key);
            return root;
        }

        if (root->val == key) {
            if (!root->left && !root->right) {
                return nullptr;
            }
            
            if (!root->left) {
                return root->right;
            }

            if (!root->right) {
                return root->left;
            }

            TreeNode* successor = root->right;
            while (successor->left) {
                successor = successor->left;
            }
            root->right = deleteNode(root->right, successor->val);
            root->val = successor->val;
            return root;
        }
        return root;
    }
};

/**

* Definition for a binary tree node.

* struct TreeNode {

* int val;

* TreeNode *left;

* TreeNode *right;

* TreeNode() : val(0), left(nullptr), right(nullptr) {}

* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}

* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

* };

class Solution {

public:

TreeNode* deleteNode(TreeNode* root, int key) {

if (!root) {

return nullptr;

}

if (root->val > key) {

root->left = deleteNode(root->left, key);

return root;

}

if (root->val < key) {

root->right = deleteNode(root->right, key);

return root;

}

if (root->val == key) {

if (!root->left && !root->right) {

return nullptr;

}

if (!root->left) {

return root->right;

}

if (!root->right) {

return root->left;

}

TreeNode* successor = root->right;

while (successor->left) {

successor = successor->left;

}

root->right = deleteNode(root->right, successor->val);

root->val = successor->val;

return root;

}

return root;

}

};