Question

我正在研究C中的数据结构，我需要找到BST中值的中位数并删除中位数，然后展示新的BST但我没有管理。我真的不知道哪里出错了请帮我解决这个问题。我还需要能够找到最接近均值的值并将其删除

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

struct Node
{
int key;
struct Node* left, *right;
};

//Function to create new BST NODE
struct Node *newNode (int item)
{
struct Node *temp = (struct Node *)malloc(sizeof(struct Node));
temp->key = item;
temp->left = temp->right = NULL;
return temp;
}
//Function to perform INORDER traversal operation
void inorder(struct Node *root)
{
if(root!=NULL)
{
inorder(root->left);
printf("%d\n",root->key);
inorder (root->right);
}
} 
//Function to insert new node with given key in BST
struct Node* insert (struct Node* node, int key)
{
//if tree is empty
if(node==NULL) return newNode(key);
//else, recur down the tree
if(key< node->key)
node->left  = insert(node->left,key);
else if(key > node->key)
node->right  = insert(node->right,key);
//Return the (unchanged) node pointer
return node;    
}


//Given a non-empty BST ,return node with minimum key value found in that          tree. Whole tree not really searched
struct Node * minValueNode(struct Node* node)
{
struct Node* current =node;
//loop down to find the left-most leaf
while (current->left !=NULL)
    current = current->left;
return current;
}
//Given the BST and key, function to deleted the key to return new root
struct Node* deleteNode(struct Node* root, int key)
{
if(root == NULL) return root;
if(key<root->key)
    root->left = deleteNode(root->left,key);
else if (key> root->key)
    root->right = deleteNode(root->right, key);
else
{
    //node with one or no child
    if(root->left == NULL)
    {
        struct Node *temp = root->right;
        free(root);
        return temp;
    }
 else if (root->right == NULL)
 {
    struct Node *temp = root->left;
    free(root);
    return temp;
}
//node with two children
struct Node* temp = minValueNode(root->right);
//copy the inorder successor to this node
root->key = temp->key;
//delete the inorder successor
root->right = deleteNode(root->right, temp->key);
}
return root;
}
//Function to count nodes in a BST using Morris Inorder traversal
int counNodes(struct Node *root)
{
struct Node *current, *pre;
int count =0;
if(root == NULL)
    return count;
current = root;
while(current!=NULL)
{
    if(current->left == NULL)
    {
        count++;
        current = current->right;;
    }
    else
    {
        pre = current->left;
        while(pre->right !=NULL && pre->right !=current) 
        pre=pre->right;
        if(pre->right == NULL)
        {
            pre->right = current;
            current= current->left;
        }
        else
        {
            pre->right = NULL;
            count++;
            current = current->right;
        }
    }
}
return count;
} 
/*FUNCTION TO FIND MEDIAN */
int FINDM(struct Node *root)
{
if(root == NULL)
    return 0;
    int count = counNodes(root);
    int currCount=0;
    struct Node *current = root, *pre, *prev;
    while(current !=NULL)
    {
        if(current->left == NULL)
        {
            currCount++;
            if(count%2!=0 && currCount ==(count+1)/2)
                return (prev->key);
                //EVEN NUMBERS
                else if(count%2==0 && currCount==(count/2)+1)
                    return (prev->key + current->key)/2;
                    prev = current;
                    current = current->right;
                }
                else
                {
                    pre=current->left;
                    while(pre->right !=NULL && pre->right !=current)
                    pre= pre->right;
                    if(pre->right = NULL)
                    {
                        pre->right = current;
                        current = current->left;
                    }
                    else
                    {
                        pre->right = NULL;
                        prev = pre;
                        currCount++;
                        if(count%2!=0 && currCount == (count+1)/2)
                            return (current->key);
    else if(count%2==0 && currCount ==  (count/2)+1)
                                return (prev->key+current->key)/2;
                                prev = current;
                                current = current->right;
                            }
                        }
                    }
                }

  //Driver Program to test functions above
int main()
{
// values are 28,90,0,32,56,78,212,34,62
struct Node *root =NULL;

root=insert(root,65);
insert(root, 28);
insert(root, 90);
insert(root,0);
insert(root, 32);
insert(root,56);
insert(root, 78);
insert(root,212);
insert(root,34);
insert(root,62);

printf("InOrder Traversal of Tree\n");
inorder(root);

printf("\nDelete 65!\n");
root=deleteNode(root,65);
printf("Modified Tree\n");
inorder(root);

printf("\nMEDIAN ");
FINDM(root);
return 0;
}

Answer 1

您已经有一个工作函数struct Node *findNth(struct Node* root, int nth) { struct Node *current, *pre; int count = 0; if (root == NULL) return count; current = root; while (current != NULL) { if (current->left == NULL) { count++; if (count == nth) { return current; } else { current = current->right;; } } else { pre = current->left; while (pre->right != NULL && pre->right != current) pre = pre->right; if (pre->right == NULL) { pre->right = current; current = current->left; } else { pre->right = NULL; count++; if (count == nth) { return current; } else { current = current->right; } } } } printf("Could not find nth(%d)\n", nth); exit(1); } int FINDM(struct Node *root) { if (root == NULL) return 0; int count = counNodes(root); struct Node *n = findNth(root, (count + 1) / 2); return n->key; }，它按顺序遍历节点，用于计算节点数。如果在查找树中第N个节点的新函数中使用该确切代码，该怎么办：

{{1}}

在C二进制搜索树中查找并删除中位数

1 个答案: