You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.
Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.
Given the following multilevel doubly linked list:
We should return the following flattened doubly linked list:
// Iteration
Node* flatten(Node* head) { // time: O(n); space: O(1)
if (!head) return nullptr;
Node* p = head;
while (p) {
// no child
if (!p->child) {
p = p->next;
continue;
}
// has child
Node* tmp = p->child;
// connect the tail of child to the next p
while (tmp->next) {
tmp = tmp->next;
}
tmp->next = p->next;
if (p->next) p->next->prev = tmp;
// connect p to p->child and remove p->child
p->next = p->child;
p->child->prev = p;
p->child = nullptr;
}
return head;
}
// Iteration
Node* flatten(Node* head) { // time: O(n); space: O(1)
Node* cur = head;
while (cur) {
if (cur->child) {
Node* next = cur->next;
Node* last = cur->child;
while (last->next) {
last = last->next;
}
cur->next = cur->child;
cur->next->prev = cur;
cur->child = nullptr;
last->next = next;
if (next) next->prev = last;
}
cur = cur->next;
}
return head;
}
// Recursion
Node* flatten(Node* head) { // time: O(n); space: O(n)
Node* cur = head;
while (cur) {
if (cur->child) {
Node* next = cur->next;
cur->child = flatten(cur->child);
Node* last = cur->child;
while (last->next) last = last->next;
cur->next = cur->child;
cur->next->prev = cur;
cur->child = nullptr;
last->next = next;
if (next) next->prev = last;
}
cur = cur->next;
}
return head;
}
