我正在尝试编写一种将键值对添加到三叉树的方法,但显然我做错了什么,因为每当我到达标记的代码时都会出现段错误
void Tree::add(int k, Node *&r)
{
cout<<"add"<<endl;
if(r==NULL){
r = new Node(k);
//check(heap area);
}
开始问题代码
else if(r->keyCount == 1){
cout<<"adding second key";
if(r->getKey() < k){
Node * temp = new Node(r->getKey(),k,r->data[0],0);
delete r;
r = temp;
r->keyCount++;
cout<<"test"<<endl;
}
else
{
Node * temp = new Node(k,r->getKey(),0,r->data[0]);
delete r;
r = temp;
r->keyCount++;
cout<<"test"<<endl;
}
结束代码
}
else if(k < r->getKey())
{
cout<<"left"<<endl;
add(k,r->child[Node::L]);
}
else if(r->keyCount > 1 && k < r->getKey(1))
{
cout<<"middle"<<endl;
add(k,r->child[Node::M]);
}
else if(r->keyCount > 1 && k > r->getKey(1))
{
cout<<"right"<<endl;
add(k,r->child[Node::R]);
}
else
r = new Node(k);
}
我想要做的是,如果在这个特定节点中使用的 2 个密钥中只有 1 个,则用一个新节点替换当前节点,该节点在适当的位置有密钥(key[ 中的较小 val 0],key[1] 中的更大 val) 我该如何正确执行此操作?
我的代码显然删除了旧节点的地址和指针,但没有正确地将指针重新分配给新节点。
编辑 更新的代码。输出如下:
% p4
Enter pairs consisting of an int and a double. I create a
ternary tree, keeping the data in order, by int. Finish entering
data by pressing ^d
2 2
add
Entering the pair: 2, 2
1 1
add
adding second key to current node
test
Entering the pair: 1, 1
-1 -1
add
left
add
Entering the pair: -1, -1
3 3
add
right
Segmentation Fault
编辑 2 如果您想查看所有代码,这里有一个指向包含整个项目的 zip 的链接:http: //sdrv.ms/WSrLfv
编辑 3 更多错误数据 - 崩溃时 gdb 的输出
Program received signal SIGSEGV, Segmentation fault.
0x08051628 in getData (x=@0x8047554) at testTree.cc:26
26 x[k]=d;
Current language: auto; currently c++
编辑 4 逐步通过 gdb 到段错误:
Breakpoint 1, Tree::add (this=0x8047554, k=3, r=@0x8047554) at tree.cc:58
58 cout<<"add"<<endl;
(gdb) n
add
61 if(r==NULL){
(gdb) n
65 else if(r->keyCount == 1){
(gdb) n
87 else if(k < r->getKey())
(gdb) n
92 else if(r->keyCount > 1 && k < r->getKey(1))
(gdb) n
97 else if(r->keyCount > 1 && k > r->getKey(1))
(gdb) n
99 cout<<"right"<<endl;
(gdb) n
right
100 add(k,r->child[Node::R]);
(gdb) n
Breakpoint 1, Tree::add (this=0x8047554, k=3, r=@0x806416c) at tree.cc:58
58 cout<<"add"<<endl;
(gdb) n
add
61 if(r==NULL){
(gdb) n
62 r = new Node(k);
(gdb) n
107 }
(gdb) n
107 }
(gdb) n
Tree::operator[] (this=0x8047554, index=3) at tree.cc:47
47 return *(locate(index,root)->data);
(gdb) n
48 }
(gdb) n
Program received signal SIGSEGV, Segmentation fault.
0x08051628 in getData (x=@0x8047554) at testTree.cc:26
26 x[k]=d;
(gdb)