二叉树由一组称为节点的互连元素组成。起始节点是树的基础，称为根节点。每个节点通常包含一个数据项，以及指向两个子节点，左节点和右节点的指针。如果任一子节点不存在，则相应的指针为NULL。
节点可以包括计数器，该计数器记录它包含的数据项的重复数。这是一个结构的示例，它定义了存储long类型整数的节点:

typedef struct Node Node;
struct Node
{
  long item;                 // 数据项
  int count;                 // 项目的副本数量
  Node *pLeft;               // 指向左节点
  Node *pRight;              // 指向右节点
};

示例代码

#define __STDC_WANT_LIB_EXT1__ 1
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

typedef struct Node Node;

// 定义二叉树中的节点排序整数
struct Node
{
  long item;                                   // The data item
  int count;                                   // Number of copies of item
  Node *pLeft;                                 // Pointer to left node
  Node *pRight;                                // Pointer to right node
};

// 函数原型
Node *create_node(long value);                 // Create a tree node
Node *add_node(long value, Node *pNode);       // Insert a new node
void list_nodes(Node *pNode);                  // List all nodes
void free_nodes(Node *pNode);                  // Release memory

int main(void){
  long newvalue = 0;
  Node *pRoot = NULL;
  char answer = 'n';
  do
  {
    printf_s("Enter the node value: ");
    scanf_s(" %ld", &newvalue);
    if (!pRoot)
      pRoot = create_node(newvalue);
    else
      add_node(newvalue, pRoot);
    printf_s("Do you want to enter another (y or n)? ");
    scanf_s(" %c", &answer, sizeof(answer));
  } while (tolower(answer) == 'y');

  printf_s("The values in ascending sequence are:\n");
  list_nodes(pRoot);  // Output the contents of the tree
  free_nodes(pRoot);  // Release the heap memory
  return 0;
}

// 创建二叉树节点
Node *create_node(long value)
{
  Node *pNode = (Node*)malloc(sizeof(Node));
  pNode->item = value;                         // Set the value
  pNode->count = 1;                            // Set the count
  pNode->pLeft = pNode->pRight = NULL;         // No left or right nodes
  return pNode;
}

// 向树中添加新节点
Node *add_node(long value, Node *pNode)
{
  if (!pNode)                                   // If there's no node
    return create_node(value);                 // create one and return it

  if (value == pNode->item)
  {                                            // Value equals current node
    ++pNode->count;                            // so increment count 
    return pNode;                              // and return the same node
  }

  if (value < pNode->item)
  {                                            // Less than current node
    if (!pNode->pLeft)                          // and no left node
    {
      pNode->pLeft = create_node(value);       //  so create a left node
      return pNode->pLeft;                     // and return it.
    }
    else                                       // There is a left node
      return add_node(value, pNode->pLeft);    // so add value via left node
  }
  else
  {                                            // Greater than current node
    if (!pNode->pRight)
    {                                          // but no right node
      pNode->pRight = create_node(value);     // so create one
      return pNode->pRight;                   // and return it.
    }
    else                                       // There is a right node
      return add_node(value, pNode->pRight);   // so add to that.
  }
}

// 按升序列出节点值
void list_nodes(Node *pNode)
{
  if (pNode->pLeft)
    list_nodes(pNode->pLeft);

  printf_s("%10d x %10ld\n", pNode->count, pNode->item);

  if (pNode->pRight)
    list_nodes(pNode->pRight);
}

// 释放分配给节点的内存
void free_nodes(Node *pNode)
{
  if (!pNode)                       // If there's no node
    return;                       // we are done and return.

  if (pNode->pLeft)                 // If there's a left sub-tree
    free_nodes(pNode->pLeft);     // free memory for those nodes.

  if (pNode->pRight)                // If there's a right sub-tree
    free_nodes(pNode->pRight);    // free memory for those nodes.

  free(pNode);                      // Free current node memory
}

执行上面示例代码，得到以下结果:

Enter the node value: 123
Do you want to enter another (y or n)? y
Enter the node value: 234
Do you want to enter another (y or n)? y
Enter the node value: 345
Do you want to enter another (y or n)? y
Enter the node value: 456
Do you want to enter another (y or n)? n
The values in ascending sequence are:
         1 x        123
         1 x        234
         1 x        345
         1 x        456