C-二进制树：无法返回正确数目的修剪节点

Question

I need to prune a binary tree past a certain level, l and need to return the number of pruned nodes.

Here's what I got:

#include "abin.h"

int freeAB (ABin a) {
    int count = 0;
    if(a == NULL) return count;
    count = count + freeAB(a->esq);
    count = count + freeAB(a->dir);
    free(a);
    count++;
    return count;
}


int pruneAB (ABin *a, int l) {

    int count = 0;
    if(l == 0){
        count = count + freeAB((*a)->esq);
        count = count + freeAB((*a)->dir);
        (*a) = NULL;
    }
    else{
        count = count + pruneAB(&((*a)->esq), l-1);
        count = count + pruneAB(&((*a)->dir), l-1);
    }
    return count;
 }

ABIN.H:

#include <stdio.h>
#include <stdlib.h>

typedef struct lligada {
    int valor;
    struct lligada *prox;
} *LInt;

typedef struct nodo {
    int valor;
    struct nodo *esq, *dir;
} *ABin;

int pruneAB (ABin *a, int l);     

This is the output of what I should have got and what I got:

Input: (depth=2)
               8
       4              12
   2       6      10      14
 1   3   5   7   9  11  13  15

Output:
[expected] res=12
   8
 4  12

[obtained] res=8
   8
 4  12

0/10 correct answers

Interestingly, if I create something like int r = 0; and do r++; every time the if(l == 0) statement is true, and then do a print statement, it prints r 4 times.

If I added 4 to the final count I would get the correct answer. I assume that I should then add to count the number of times the if(l == 0) is true.

(I can't do it. If I do count++ I get segmentation fault)

How would you do it? Thanks.

https://codeboard.io/projects/16275

Answer 1

int pruneAB (ABin *a, int l) {
    int count = 0;
    if (!*a) return 0;
    if (l < 0) return count;
    if(l == 0){
        count = freeAB(*a);
        (*a) = NULL;
    }
    else{
        count = count + pruneAB(&((*a)->esq), l-1);
        count = count + pruneAB(&((*a)->dir), l-1);
    }
    return count;
}