如何统计二叉树的总节点数

Question

我需要计算二叉树中的节点总数。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

您声明了 c 但没有在任何地方初始化，也没有在任何地方使用。然后你打印 c 的值，它给你垃圾值。

您可以将 count(node *tree) 函数修复为

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;
    }
}

加入main

int main()
{
    .............
    .............


    c = count(root);        //number of node assign to c
    printf("Number of node %d \n",c);
}

Answer 2

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);
    }
}

Answer 3

count 函数将是

int count(struct node *root)
{
    int a=1;

    if (root==NULL){
        return 0;
    }

    else
    {
        a += count(root->left);
        a += count(root->right);
        return a;
    }
}

在main函数中，count函数的调用是这样的

printf("total no of nodes  of binary tree is %d\n",count(p));//p is root node

如何统计二叉树的总节点数

How to count the total number of nodes in binary tree

c

binary-tree

visual-studio-2010

nodes