英文:
Initialize a tree in inorder?
问题
以下是您要翻译的内容：
tree::tree(const initializer_list<int>& I) {
const int* p1 = I.begin();
root = nullptr;
IL_helper(I, p1, root);
}
void tree::IL_helper(const initializer_list<int>& I, const int* &p1, node* &p2) {
if (p1 == I.end()) {
return;
}
p2 = new node(*p1);
++p1;
IL_helper(I, p1, p2->Lchild);
IL_helper(I, p1, p2->Rchild);
}
如何使用递归的方式根据中序遍历创建一棵树。在我的代码中，我意识到它只会创建左子树。
tree t{1,2,3,4,5,6,7};
cout << t.inOrderT(t.root);
输出：1,2,3,4,5,6,7
英文:
Im using initializer list and helper to create a tree in inorder. for example {1,2,3,4,5,6,7} if I print the tree in inorder traversal it will also give 1,2,3,4,5,6,7.
tree::tree(const initializer_list<int>& I) {
const int* p1 = I.begin();
root = nullptr;
IL_helper(I, p1, root);
}
void tree::IL_helper(const initializer_list<int>& I, const int*& p1, node* & p2) {
if (p1 == I.end()) {
return;
}
p2 = new node(*p1);
++p1;
IL_helper(I, p1, p2->Lchild);
IL_helper(I, p1, p2->Rchild);
}
how to form a tree with inorder traversal with recursion. In my code, I realize it will only have Lchild.
tree t{1,2,3,4,5,6,7} ;
cout << t.inOrderT(t.root);
output: 1,2,3,4,5,6,7
答案1
得分: 0
p2 = new node(*p1); 操作会覆盖 p2，但它的类型只是 node*，所以这个赋值不会影响函数外的任何内容。只需将参数更改为 node* &p2。
另外，我注意到第二个问题 - 在你的示例中没有对输入列表进行二分，所以你构建的树节点将始终只有 Lchild 设置。因为在递归链中的 IL_helper(I, p1, p2->Lchild); 会消耗所有的列表，不会留下任何东西供 Rchild 调用。从技术上讲，如果这是预期的行为（例如，之后可能会重新平衡树），那么这不算是错误，但是吗？
英文:
p2 = new node(*p1);
overwrites the p2, but its type is just node*, so this assignment does not affect anything outside of the function. Just change the parameter to node* &p2.
Also, I see a second problem - in your example is no bisection of the input list - so the nodes of the tree you will construct will always have just the Lchild set. Because in the recursive chain of IL_helper(I, p1, p2->Lchild); all of the list will be consumed, leaving nothing left for the Rchild call. It's technically not a bug if that's the intended behavior ( for example, the tree might be re-balanced afterwards ), but is it ?
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