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二叉树转换成双向链表(day6)

admin 11月 12, 2020 0

题目

题目：输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表。
要求不能创建任何新的结点，只调整指针的指向。比如将二元查找树

将
     10
     / 
    6   14
  /   /  
 4  8 12  16
 转换成双向链表4=6=8=10=12=14=16

实现

struct Node {
    int value;
    Node* left;
    Node* right;
};
bool (Node* head, Node* &left, Node* &right){
    left = head;
    right = head;
    if(!head){
        return false;
    } else {
        Node *l1, *r1, *l2, *r2;
        l1 = NULL;
        r1 = NULL;
        l2 = NULL;
        r2 = NULL;
        if(createList(head->left, l1, r1)){
            head->left = r1;
            r1->right = head;
            left = l1;
        }
        if(createList(head->right, l2, r2)){
            head->right=l2;
            l2->left=head;
            right = r2;
        }
        return true;
    }
}
int main(){
    Node n4={4,NULL,NULL};
    Node n5={8,NULL,NULL};
    Node n6={12,NULL,NULL};
    Node n7={16, NULL, NULL};
    Node n2={6,&n4,&n5};
    Node n3={14,&n6,&n7};
    Node n1={10,&n2,&n3};
    Node*Left=NULL;
    Node*Right=NULL;
    createList(&n1,Left,Right);
    while(Left){
        cout<<Left->value<<endl;
        Left=Left->right;
    }
    return 0;
}