删除二叉树的功能

时间:2015-02-12 02:46:22

标签: c function struct tree

鉴于以下代码,我可以使用什么来制作删除功能?我尝试了很多东西,并且一直试图让它发挥作用。我的主要问题是尝试删除具有左右子节点的节点。对于没有子节点的节点,我可以将其父节点设置为指向null并释放节点。对于一个孩子,只需将父项设置为指向子项并释放该节点即可。如何在概念上和我的代码中为具有两个孩子的节点执行此操作?

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

struct bin_tree {
  int data;
  struct bin_tree * right, * left;
} bin_tree;
typedef struct bin_tree node;
void help()//help
{
  printf("Options:\n");
  printf(" # -Put in any number to add it to the tree if not already there\n");
  printf(" s # -Put in s and a number to search the tree for the number\n");
  printf(" d # -Delete the number from the tree\n");
  printf(" p -Put in p to print the tree\n");
  printf(" ? -At any time you can press ? to display the help message\n");
  printf(" Q -If you decide the leave the realm of the tree then you can press Q to quit this program\n");

}

int max(int a,int b)//max tree length
{
  if(a>b)
    return a;
  else
    return b;
}

int height(node* tree)//height
{
  if(tree != NULL)
    return(1 + max(height(tree->left),height(tree->right)));
  else
    return 0;
}

void insert(node ** tree, int val)//insert
{
  node *temp = NULL;
  if(!(*tree))
  {
    temp = (node *)malloc(sizeof(node));
    temp->left = temp->right = NULL;
    temp->data = val;
    *tree = temp;
    return;
  }

  if(val < (*tree)->data)
  {
    insert(&(*tree)->left, val);
  }
  else if(val > (*tree)->data)
  {
    insert(&(*tree)->right, val);
  }
}
void print(node * tree)//print
{
  if (tree)
  {
    print(tree->left);
    printf("[%d] ",tree->data);
    print(tree->right);
  }
}
node* search(node ** tree, int val)
{//search
  if(!(*tree))
  {
    return NULL;
  }
  if(val < (*tree)->data)
  {
    search(&((*tree)->left), val);
  }
  else if(val > (*tree)->data)
  {
    search(&((*tree)->right), val);
  }
  else if(val == (*tree)->data)
  {
    return *tree;
  }
}

void main()
{
  node *root;
  node *tmp;
  int no;
  char ch, buff[500];

  root = NULL;
  printf("Options:\n");
  printf(" # -Put in any intiger to add it to the tree if not already there\n");
  printf(" s # -Put in s and a number to search the tree for the number\n");
  printf(" d # -Delete the number from the tree\n");
  printf(" p -Print the tree\n");
  printf(" ? -At any time you can press ? to display the help message\n");
  printf(" Q -If you decide the leave the realm of the tree then you can press Q to quit this program\n");
  while(1){
    printf(">");
    fgets(buff,499,stdin); //grabs input from user
    if(sscanf(buff,"%i",&no)==1){//decides if just a number
      tmp = search(&root, no);//looks for number in the tree
      if (tmp)
      {
        printf("Node already in tree!\n", tmp->data);
      }
      else 
      {
        insert(&root, no);//if not in tree insert it
      }
    }
    else if(sscanf(buff,"%c %i",&ch,&no)>=1)//checks if character
    {
      switch(ch)
      {
        case 's'://search for number
        {
          tmp = search(&root, no);
          if (tmp)
          {
            printf("Node found=%d\n", tmp->data);
          }
          else
          {
            printf("Node not found in tree.\n");
          }
          break;
        }
        case 'd':
          tmp = search(&root, no);
          if (tmp)
          {
            //Call delete function
            printf("Node %i deleted", no);
            break;
          }
          else
          {
            printf("Node not found in tree.\n");
            break;
          }
        case 'Q'://quit
          exit(0);
        case 'p'://print tree
          printf("\n\n");
          print(root);
          printf("\nHeight= %i\n\n",height(root));
          break;
        case '?'://display help
          help();
          break;
        default://idiot >.>
          printf("Invalid input!\n\n");
          help();
          break;
      }
    }
  }
  return;
}

1 个答案:

答案 0 :(得分:2)

左边的最大节点或右边的最小节点都会占据它的位置!

只需将其中一个放在已删除节点的位置(并且delete来自您之前的位置),您的树仍然是有效的二叉搜索树。看一下这个例子:

    15
   /  \
  …    25
      /  \
     20  30
       \
        23

假设您要删除节点25

  • 通过二叉搜索树的属性,您已经知道所有子节点都必须大于父节点(15),因此使用其中一个而不是25是有效的。 ✓
  • 如果从左子树(23)中选择最大的节点,它将比左边的任何节点都大,但它也会小于右边的任何节点,因此它很适合中间,可以取代已删除的节点。 ✓
  • 右子树(30)中的最小节点也是如此。 ✓

如果所选节点是叶子,一切都很好,你可以删除它。否则,在您选择的节点上执行delete操作。

您还可以使用look at the Wikipedia article of the binary search tree进行伪代码实现。