450. Delete Node in a BST

最新推荐文章于 2025-04-02 00:39:14 发布

Aurora2005

最新推荐文章于 2025-04-02 00:39:14 发布

阅读量784

点赞数 16

分类专栏：代码随想录数据结构 leetcode 文章标签：算法

本文链接：https://blog.youkuaiyun.com/2301_80161204/article/details/143116312

版权

数据结构同时被 3 个专栏收录

122 篇文章

订阅专栏

leetcode

116 篇文章

订阅专栏

代码随想录

77 篇文章

订阅专栏

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.

Example 1:

Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: 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 above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

Example 2:

Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.

Example 3:

Input: root = [], key = 0
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 104].
-105 <= Node.val <= 105
Each node has a unique value.
root is a valid binary search tree.
-105 <= key <= 105

Follow up: Could you solve it with time complexity O(height of tree)?

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        //第一种情况，没有找到要删除的节点
        if(root==NULL)return root;
        if(root->val==key){
            //第二种情况：找到要删除的节点，其左右孩子都为空
            if(root->left==NULL && root->right==NULL){
                delete root;
                return NULL;
            }
            //第三种情况：左空，右不空
            else if(root->left==NULL && root->right!=NULL){
                auto node=root->right;
                delete root;
                return node;
            }
            //第四种情况：左不空，右空
            else if(root->left!=NULL && root->right==NULL){
                auto node=root->left;
                delete root;
                return node;
            }
            //第五种情况：左右都不空
            else{
                TreeNode*node=root->right;
                while(node->left!=NULL){
                    node=node->left;
                }
                node->left=root->left;
                TreeNode*temp=root;
                root=root->right;
                delete temp;
                return root;
            }
        }
        if(root->val>key)root->left=deleteNode(root->left,key);
        if(root->val<key)root->right=deleteNode(root->right,key);
        return root;
    }
};