Delete Node in a BST

最新推荐文章于 2020-01-24 14:15:48 发布

无名_1989

最新推荐文章于 2020-01-24 14:15:48 发布

阅读量94

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/dongyu_1989/article/details/80769011

本文介绍了一种在二叉搜索树(BST)中删除指定键值节点的方法，并提供了完整的C++实现代码。通过两阶段的过程——查找要删除的节点并进行删除操作——确保了树结构的正确维护。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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) {
       TreeNode * pre = NULL;
       TreeNode * cur = root;
       while(cur)
       {
           if(cur->val==key) break;
           pre = cur;
           if(cur->val>key) cur = cur->left;
           else cur = cur->right;
       }
       if(!cur) return root;
       if(!pre) return del(cur);
       if(pre->left&&pre->left->val==key) pre->left = del(cur);
       else pre->right = del(cur);
       return root;
    }
    TreeNode *del(TreeNode *node)
    {
        if(!node->left&&!node->right) return NULL;
        if(!node->left || !node->right) 
        {
            return node->left ? node->left : node->right; 
        }
        TreeNode * cur = node->right;
        TreeNode * pre = node;
        while(cur->left)
        {
            pre = cur;
            cur = cur->left ;
        }
        node->val = cur->val;
        ( node == pre ? node->right : pre->left )= cur->right ;
        return node;
    }
};