我需要计算二叉树中的节点总数。当我执行此代码时,问题就出现了,它为节点总数提供了垃圾值。我的程序输出类似993814
。应该是7
。
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);
}
答案 0 :(得分:7)
我宁愿通过在不使用局部变量的情况下在每个递归调用中返回总和来实现它。
int count(struct node *root){
if(root == NULL){
return 0;
}
else{
return 1 + count(root->left) + count(root->right);
}
}
答案 1 :(得分:4)
您声明c
但不能初始化,也不会在任何地方使用。
然后打印c
的值,它会为您提供垃圾值。
您可以将count(node *tree)
功能修复为
int count(node *tree)
{
int c = 1; //Node itself should be counted
if (tree ==NULL)
return 0;
else
{
c += count(tree->left);
c += count(tree->right);
return c;
}
}
添加main
int main()
{
.............
.............
c = count(root); //number of node assign to c
printf("Number of node %d \n",c);
}