求二叉树从叶子到根节点的值组成数字的和

题目：如下图二叉树，每个叶子节点到根节点的数值可以组成一个数字，如253/753/83，写代码求和，即253+753+83=1089

基本思路：利用栈，后序遍历二叉树，顺便求和，代码如下：

#include 
#include 
#include 

#define NUM 5

typedef struct node{
    int val;
    int l;
    int r;
}tree;

typedef struct stack{
    int top;
    tree s[NUM];
}tree_stack;
tree_stack s;

int is_empty(tree_stack *s)
{
    return (s->top == -1);
}
int is_full(tree_stack *s)
{
    return (s->top >= NUM);
}
int push(tree_stack *s, tree *n)
{
    if(is_full(s))
        return -1;
    s->top++;
    memcpy(s->s + s->top, n, sizeof(tree));

    return 0;
}
int pop(tree_stack *s, tree *n)
{
    if(is_empty(s) || !n)
        return -1;
    memcpy(n, s->s + s->top, sizeof(tree));
    s->top--;
    
    return 0;
}

void init_tree(tree *t)
{
    t[0].val = 3;
    t[0].l = 1;
    t[0].r = 2;
    
    t[1].val = 5;
    t[1].l = 3;
    t[1].r = 4;
    
    t[2].val = 8;
    t[2].l = -1;
    t[2].r = -1;
    
    
    t[3].val = 2;
    t[3].l = -1;
    t[3].r = -1;
    
    t[4].val = 7;
    t[4].l = -1;
    t[4].r = -1;
}
int calcute(tree_stack *s)
{
    tree n;
    int val,i;

    pop(s,&n);
    val = 0;
    if(n.l==-1 && n.r == -1)
    {
        val = n.val * pow(10,s->top+1);
        for(i=s->top;i>=0;i--)
        {
            val += s->s[i].val * pow(10,i);
        }
        printf("val:%d\n",val);
    }
    return val;
}
void show_tree(tree *t, int index, int *total)
{
    if(index < 0)
        return;
    push(&s,t+index);
    show_tree(t,t[index].l,total);
    show_tree(t,t[index].r,total);
    *total += calcute(&s);
}

int main()
{
    int total = 0;
    tree t[NUM];
    s.top = -1;
    init_tree(t);
    show_tree(t,0, &total);
    printf("total:%d\n",total);
    return 0;
}