int hight(struct node* root)
{
if(root==NULL)
return 0;
else
{
int l=hight(root->left);
int r=hight(root->right);
return l>r?(l+1):(r+1);
}
}
int hight(TREE T)
{
h++;
if(maxh<h) maxh=h;
if(T){
hight(T->left);
hight(T->right);
}
h--;
return maxh-1;
}
int binatTreeMain(){
TREE T;
int choice;
char c='y';
printf("请输入带空二叉树先序遍历序列（0表示空节点）,例如3个元素的二叉树输入为\n");
printf(" 1 2 0 0 3 0 0 (1为根结点)：");
creattree(&T);
while(c=='Y'||c=='y'){
choice=menu();
switch(choice){
case 11:printf("递归法前序遍历：");pretraverse1(T);break;
case 12:printf("非递归法前序遍历：");pretraverse2(T);break;
case 21:printf("递归法中序遍历：");intraverse1(T);break;
case 22:printf("非递归法中序遍历：");intraverse2(T);break;
case 3:printf("递归法后序遍历：");posttraverse1(T);break;
case 41:outdegree1(T);
printf("出度为0的节点数目：%d\n",n0);
printf("出度为1的节点数目：%d\n",n1);
printf("出度为2的节点数目：%d\n",n2);break;
case 42:outdegree2(T);break;
case 5:floortraverse(T);break;
case 6:printf("\n此二叉树高度为：%d",hight(T));break;
case 7:exchange(&T);intraverse1(T);break;
}
printf("\n是否继续？(Y/N):");
fflush(stdin);
scanf("%c",&c);
system("cls");
}
}
int maxwidth(struct node *root)
{
if(root)
{
int h=hight(root);
int *con=(int *)calloc(sizeof(int),h);
int level=0;
getwidth(root,con,level);
return ma(con,h);
}
}
