我需要计算二叉树中的节点总数。现在问题出现在我执行这段代码时,它给出了节点总数的垃圾值。我的程序的输出是993814
。应该是7
。如何计算二叉树中的节点总数
如何解决这个问题?
#include<stdlib.h>
#include<stdio.h>
struct binarytree
{
int data;
struct binarytree * right, * left;
};
typedef struct binarytree node;
void insert(node ** tree, int val)
{
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_preorder(node * tree)
{
if (tree)
{
printf("%d\n",tree->data);
print_preorder(tree->left);
print_preorder(tree->right);
}
}
void print_inorder(node * tree)
{
if (tree)
{
print_inorder(tree->left);
printf("%d\n",tree->data);
print_inorder(tree->right);
}
}
int count(node *tree)
{
int c=0;
if (tree ==NULL)
return 0;
else
{
c += count(tree->left);
c += count(tree->right);
return c;
}
}
void print_postorder(node * tree)
{
if (tree)
{
print_postorder(tree->left);
print_postorder(tree->right);
printf("%d\n",tree->data);
}
}
int main()
{
node *root;
node *tmp;
int c;
root = NULL;
/* Inserting nodes into tree */
insert(&root, 9);
insert(&root, 10);
insert(&root, 15);
insert(&root, 6);
insert(&root, 12);
insert(&root, 17);
insert(&root, 2);
/* Printing nodes of tree */
printf("Pre Order Display\n");
print_preorder(root);
printf("In Order Display\n");
print_inorder(root);
printf("Post Order Display\n");
print_postorder(root);
printf("Number of node %d \n",c);
}
如果通过在insert()中递增和在remove()上递减来追踪内部计数,那么最简单。 –