c - 如何统计二叉树的总节点数

Question

我需要计算二叉树中节点的总数。当我执行这段代码时，现在出现了问题，它为节点总数提供了垃圾值。我程序的输出类似于 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);

}

Answer 1

我宁愿在不使用局部变量的情况下在每次递归调用中返回总和。

int count(struct node *root){
    if(root == NULL){
        return 0;
    }
    else{
        return 1 + count(root->left) + count(root->right);
    }
}