Question

当我运行此代码以获取二叉树的长度时，我得到一个分段错误：11错误我已经尝试纠正它，我可以让它运行的唯一方法是通过调用大小函数只为左或右节点。当我以这种方式运行它（根据我的说法是正确的）我得到了错误。

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

struct node {
    int data;
    struct node* left;
struct node* right;
};

typedef struct node node;

node* newNode( int data ){
    node* node = malloc( sizeof(node) );
    assert(node);
    node->data = data;
    node->left = NULL;
    node->right = NULL;

    return node;
}

node* insert( node* node, int data ) {
    if( node == NULL){
        return newNode(data);
    }
    else{
        if( data <= node->data ){
            node->left = insert(node->left, data);
        }
        else{
            node->right = insert(node->right,data);
        }
    }
    return node;
}

node* buildOneTwoThree() {
    node* root = newNode(2);
    root->left = newNode(1);
    root->right = newNode(5);
    return root;
}

int size( node* tree ) {
    int n = 0;
    if( tree == NULL ){
        return 0;
    } else {
        return size(tree->left) + 1 + size(tree->right);
    }
}


int main(){

    node* tree = NULL;
    tree = buildOneTwoThree();
    printf("size = %i \n", size(tree)+size(tree->right) );

    return 0;
}

Answer 1

更改

node* node = malloc( sizeof(node) );//<<-sizeof(node) : It has been interpreted as a variable name, not the type.

到

node* node = malloc( sizeof(struct node) );

分段故障11在C中构建二叉树

1 个答案: