拷贝二叉树

拷贝二叉树，就是要拷贝根节点，叶子节点，第一步得先分配一个空间给根节点， BiTNode* newnode = (BiTNode*)malloc(sizeof(BiTNode));将其叶子节点指向NULL，在判断原树的根节点的左节点和右节点是否为空，不为空就得复制过去，利用迭代就很容易做到。最后将新的子节点和根节点链接起来。

typedef struct BiTNode
{
int data;
struct BiTNode* lchild, *rchild;
}BiTNode;

BiTNode* CopyTree(BiTNode *root)
{
   if (root == NULL)
   {
       return NULL;
   }
   BiTNode* newnode = (BiTNode*)malloc(sizeof(BiTNode));
   BiTNode* newlc = NULL;
   BiTNode* newrc = NULL;
   newnode->data = root->data;//拷贝根节点
   if (root->lchild != NULL)
   {
       //拷贝左子树
       newlc = CopyTree(root->lchild);
   }
   else
   {
       newlc = NULL;
   }
   if (root->rchild != NULL)
   {
       //拷贝右子树
       newrc = CopyTree(root->rchild);
   }
   else
   {
       newrc = NULL;
   }
   newnode->lchild = newlc;
   newnode->rchild = newrc;
   return newnode;
}
void main()
{
   BiTNode t1, t2, t3, t4, t5;
   memset(&t1, 0, sizeof(BiTNode));
   memset(&t2, 0, sizeof(BiTNode));
   memset(&t3, 0, sizeof(BiTNode));
   memset(&t4, 0, sizeof(BiTNode));
   memset(&t5, 0, sizeof(BiTNode));

   t1.data = 1;
   t2.data = 2;
   t3.data = 3;
   t4.data = 4;
   t5.data = 5;

   t1.lchild = &t2;
   t1.rchild = &t3;
   t2.rchild = &t4;
   t3.lchild = &t5;
   inOrder(&t1);
   printf("\n");
   BiTNode *newn=CopyTree(&t1);
   inOrder(newn);
   system("pause");

}

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