BST - 删除节点

Question

所以我正在尝试解决log2base2.com 中涉及从 BST 中删除节点的问题。除非尝试删除具有两个子节点的节点，否则它可以工作。我知道我可以简单地替换数字，然后在最后删除重复项，但是如果我的结构中的数据不仅仅是key 怎么办？那请问我做错了什么？

#include<stdio.h>
#include<stdlib.h>

struct node
{
    int key;
    struct node *left;
    struct node *right;
};

int getRightMin(struct node *root)
{
    //Write your code here
    struct node *temp = root;
    
    while(temp->left!=NULL)
    {
        temp=temp->left;
    }
    
    return temp;
    
}

struct node *removeNode(struct node *root, int key)
{   
    //Write your code here
    if(root == NULL){return NULL;}
    if(key<root->key){root->left = removeNode(root->left, key);}
    else if(key>root->key){root->right = removeNode(root->right,key);}
    else
    {
        if(root->left==NULL && root->right==NULL)
        {
            free(root);
            return NULL;
        }
        else if(root->left)
        {
            struct node *temp = root->left;
            free(root);
            return temp;
        }
        else if(root->right)
        {
            struct node *temp = root->right;
            free(root);
            return temp;
        }
        else
        {
            struct node *temp = getRightMin(root->right);
            temp->left = root->left;
            temp->right = root->right;
            root = temp;
            root->right = removeNode(root->right, temp->key);
        }
    }
    
    return root;
}

//Don't change the below code
struct node *getNewNode(int val)
{
    struct node *newNode = malloc(sizeof(struct node));
    newNode->key   = val;
    newNode->left  = NULL;
    newNode->right = NULL;

    return newNode;
}

struct node *insert(struct node *root, int val)
{
    if(root == NULL)
        return getNewNode(val);
    if(root->key < val)
        root->right = insert(root->right,val);
    else if(root->key > val)
        root->left = insert(root->left,val);

    return root;
}

void inorder(struct node *root)
{
    if(root == NULL)
        return;
    inorder(root->left);
    printf("%d ",root->key);
    inorder(root->right);
}

int main()
{
    struct node *root = NULL;
    root = insert(root,100);
    root = insert(root,50);
    root = insert(root,200);
    root = insert(root,150);
    root = insert(root,300);

    int key;
    scanf("%d",&key);

    root = removeNode(root,key);
    inorder(root);

    return 0;
}

测试用例 1：

输入：200 预期输出：50 100 150 300 实际输出：50 100 150

测试用例 2：

输入：100 预期输出：50 150 200 300 实际输出：50

Answer 1

您必须将removeNode() 中的条件重新排序为：

        if(root->left==NULL && root->right==NULL)
        {
            free(root);
            return NULL;
        }
        else if (root->left && root->right)
        {
            struct node *temp = getRightMin(root->right);
            temp->left = root->left;
            temp->right = root->right;
            root = temp;
            root->right = removeNode(root->right, temp->key);
        }
        else if(root->left)
        {
            struct node *temp = root->left;
            free(root);
            return temp;
        }
        else // root->right != NULL
        {
            struct node *temp = root->right;
            free(root);
            return temp;
        }

看起来(root->left && root->right) 的案件被(root->left) 抓住了，并且 对root->right 的引用丢失。