用非递归算法来数树的节点(求看看，请了！！)

Question

#include "stdio.h"
#include<stdlib.h>
typedef struct Ttree
{int data;
struct Ttreelchild;
struct Ttreerchild;
}tree;

typedef struct stack
{tree *data[100];
int top;
}stack;

tree*create(tree *T)
{T->data=1;
tree n1=NULL;
tree n2=NULL;
n1=(tree)malloc(sizeof(tree));
n2=(tree)malloc(sizeof(tree));

n1->data=2;
n2->data=3;
T->lchild=n1;
T->rchild=n2;
tree n3=NULL;
n3=(tree)malloc(sizeof(tree));

n3->data=4;
n1->lchild=n3;
tree n4=NULL;
n4=(tree)malloc(sizeof(tree));

n4->data=5;
n1->rchild=n4;
tree n5=NULL;
n5=(tree)malloc(sizeof(tree));

n5->data=6;
n2->lchild=n5;
tree n6=NULL;
n6=(tree)malloc(sizeof(tree));

n6->data=7;
n2->rchild=n6;
tree n7=NULL;
n7=(tree)malloc(sizeof(tree));

n7->data=8;
n6->lchild=n7;
tree n8=NULL;
n8=(tree)malloc(sizeof(tree));

n8->data=9;
n2->rchild=n8;
return T;
}

void initstack(stack *L)
{L->top=-1;
}

tree*count(tree *T,stack *L)
{tree *a[100];
int h,i;
h=0;
i=0;
tree p=NULL;
p=(tree)malloc(sizeof(tree));
p=T;
L->top++;
L->data[L->top]=p;
while(L->top!=-1)
{
while(p->lchild!=NULL)
{L->top++;
L->data[L->top]=p->lchild;
p=p->lchild;
}
while(p!=T&&L->top!=-1)
{if(p->lchild==NULL)
{ if(p->lchild==NULL)
{
a[i++]=L->data[L->top];
L->top--;
h++;}
else {
L->top++;
L->data[L->top]=p->lchild;
p=p->lchild;
}}
if(L->top!=-1)
{if(p->rchild==NULL&&L->top!=-1)
{
p=L->data[L->top];
a[i++]=L->data[L->top];
L->top--;
h++;
}
else {p=p->rchild;
L->top++;
L->data[L->top]=p;
while(p->lchild!=NULL)
{L->top++;
L->data[L->top]=p->lchild;
p=p->lchild;
}
}
}

}

p=p->rchild;
L->top++;
L->data[L->top]=p;
}
T->data=h;
return T;
}

int main()
{int h;
stack L;
tree T;
T=(tree)malloc(sizeof(tree));

T=create(T);

initstack(&L);
T=count(T,&L);
printf("输出二叉树结点的个数：%d",T->data);
return 0;
}

Answer 1

你这个有点乱，这是我写的你可以参考一下，进行中序遍历后输出节点个数：

#include <stdio.h>
int cnt=0;
typedef struct node{ //树的结点
    int data;
    struct node* left;
    struct node* right;
} Node;

typedef struct { //树根
    Node* root;
} Tree;

void insert(Tree* tree, int value)//创建树
{
    Node* node=(Node*)malloc(sizeof(Node));//创建一个节点
    node->data = value;
    node->left = NULL;
    node->right = NULL;
    if (tree->root == NULL)//判断树是不是空树
    {
        tree->root = node;
    }
    else {//不是空树
        Node* temp = tree->root;//从树根开始
        while (temp != NULL)
        {


            if (value < temp->data)//小于就进左儿子
            {
                if (temp->left == NULL)
                {
                    temp->left = node;
                    return;
                }
                else {//继续判断
                    temp = temp->left;
                }
            }
            else {//否则进右儿子

                if (temp->right == NULL)
                {
                    temp->right = node;
                    return;
                }
                else {//继续判断
                    temp = temp->right;
                }
            }
        }
    }
    return;
}

void inorder(Node* node)//树的中序遍历
{
    if (node != NULL)
    {
        inorder(node->left);
        printf("%d ",node->data);
        cnt++;
        inorder(node->right);
    }
}

int main()
{
    Tree tree;
    tree.root = NULL;//创建一个空树
    int n;
    scanf("%d",&n);
    for (int i = 0; i < n; i++)//输入n个数并创建这个树
    {
        int temp;
        scanf("%d",&temp);
        insert(&tree, temp);
    }
    printf("中序遍历的结果是：");

    inorder(tree.root);//中序遍历
    printf("\n一共有%d个节点",cnt);
    getchar(); getchar();
    return 0;
}