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关于二叉树的问题

时间:2011-12-16

来源:互联网

struct tree
{
int data;
struct tree *lchild,*rchild;
}t1,t2,t3,t4,t5,t6,t7;


main()
{
struct tree t1,t2,t3,t4,t5,t6,t7;
t1.data=1;
t1.lchild=0;
t1.rchild=0;
t2.data=2;
t2.lchild=t1;
t2.rchild=t4;
t3.data=3;
t3.lchild=0;
t3.rchild=0;
t4.data=4;
t4.lchild=t3;
t4.rchild=t5;
t5.data=5;
t5.lchild=0;
t5.rchild=0;
t6.data=6;
t6.lchild=t2;
t6.rchild=t7;
t7.data=7;
t7.lchild=0;
t7.rchild=0;
}

这样建立一个二叉树应该没问题吧??但是系统报错no operator defined which takes a right-hand operand of type 'struct tree' (or there is no acceptable conversion) 每个lchild 和rchild的赋值都这样报错???
求高手指教一下!!

作者: abc461658920   发布时间: 2011-12-16

C/C++ code
struct tree {
    int data;
    struct tree *lchild,*rchild;
};

main()
{
    struct tree t1,t2,t3,t4,t5,t6,t7;
    t1.data=1;
    t1.lchild=0;
    t1.rchild=0;
    t2.data=2;
    t2.lchild=&t1;
    t2.rchild=&t4;
    t3.data=3;
    t3.lchild=0;
    t3.rchild=0;
    t4.data=4;
    t4.lchild=&t3;
    t4.rchild=&t5;
    t5.data=5;
    t5.lchild=0;
    t5.rchild=0;
    t6.data=6;
    t6.lchild=&t2;
    t6.rchild=&t7;
    t7.data=7;
    t7.lchild=0;
    t7.rchild=0;
}

作者: yulinlang   发布时间: 2011-12-16