从堆栈创建二叉树？
我的任务是创建一个程序，将 ((X+3)*(X+4)) 之类的东西转换为二叉树，并具有一些其他功能。到目前为止，我已经接受了输入，并将其解析为两个堆栈，一个包含操作数，另一个包含运算符。
我现在只是简单地定义了堆栈（因此它们只有一个 nextnode 和 char 值。 但是，我似乎在将堆栈中的值添加到树中时遇到问题（因此可能是定义树时出现问题）。
我的堆栈定义如下：
typedef struct node
{
char value;
struct node * nextnode;
} node;
我的树定义：
typedef struct tree
{
node * thisNode;
struct tree *right, *left;
} tree;
我不确定节点*部分，也许它应该有所不同。
对于初学者来说，我一直在考虑 2+3 的简单情况。在这种情况下，树的根应该是+，左边是2，右边是3。
+
/\
2 3
如何将堆栈上的东西添加到我的树中？我尝试过使用
root->thisNode = operatorTop;
其中operatorTop是操作符堆栈的顶部（定义为node *operatorTop）， 但即使是这条简单的线似乎也存在段错误。
I'm tasked with creating a program that turns something like ((X+3)*(X+4)) into a binary tree, along with some other features. So far, I have taken in the input, and parsed it into two stacks, one containing the operands, the other the operators.
I defined the stacks simply for now (so they only have a nextnode and char value.
However, I seem to have problems adding values from the stacks into my tree (so probably a problem in defining the tree).
My stack is defined as such:
typedef struct node
{
char value;
struct node * nextnode;
} node;
My tree is defined:
typedef struct tree
{
node * thisNode;
struct tree *right, *left;
} tree;
I'm not sure about the node* part, perhaps it should be something different.
I've been considering the simple case of 2+3 for starters. In this case, the root of the tree should be +, with left being 2 and right being 3.
+
/\
2 3
How to add something that's on a stack to my tree? I have tried using
root->thisNode = operatorTop;
Where operatorTop is the top of the operator stack (defined as node * operatorTop),
but even that simple line seems to segfault.
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也许问题是您没有调用
malloc()来为root保留空间。（编译器将报告空指针分配，但在空点取消引用时会出现段错误，因为您指向随机位置。）
Maybe the problem is you have not called
malloc()to reserve the space forroot.(Compilers will report null pointer assignment, but segfault on null point dereference because you are pointing to a random place.)