[Leetcode] 116, 129, 23
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116. Populating Next Right Pointers in Each Node
Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL
.
Initially, all next pointers are set to NULL
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1 / \ 2 3 / \ / \ 4 5 6 7
After calling your function, the tree should look like:
1 -> NULL / \ 2 -> 3 -> NULL / \ / \ 4->5->6->7 -> NULL
Solution: 递归,注意题目中已经限定了是perfect binary tree ,因此不需要讨论很多种情况。
Code:
/** * Definition for binary tree with next pointer. * struct TreeLinkNode { * int val; * TreeLinkNode *left, *right, *next; * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */class Solution {public: void connect(TreeLinkNode *root) { connect(root, NULL); }private: void connect(TreeLinkNode *root, TreeLinkNode *next){ if(root==NULL) return; root->next = next; connect(root->left, root->right); connect(root->right, next!=NULL?next->left:NULL); }};
129. Sum Root to Leaf Numbers
Given a binary tree containing digits from 0-9
only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3
which represents the number 123
.
Find the total sum of all root-to-leaf numbers.
For example,
1 / \ 2 3
The root-to-leaf path 1->2
represents the number 12
.
The root-to-leaf path 1->3
represents the number 13
.
Return the sum = 12 + 13 = 25
.
Solution: 树的递归。
Code:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: int sumNumbers(TreeNode* root) { sumNumbers(root, 0); return sum; }private: int sum = 0; void sumNumbers(TreeNode* root, int cur){ if(root==NULL) return; cur = cur*10 + root->val; if(root->left==NULL && root->right==NULL){ sum += cur; }else{ sumNumbers(root->left, cur); sumNumbers(root->right, cur); } cur = cur/10; }};
23. Merge k Sorted Lists
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
Solution(1):直接做,时间复杂度为O(kn),k是数组list的长度,n是待排序的数的总数,当k比较大的时候,无法等同于O(n),Leetcode上会超时。
Code:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* mergeKLists(vector<ListNode*>& lists) { ListNode dummy(INT_MAX); ListNode* cur = &dummy; int mini = -1; int k = 20; while(1){ mini = -1; for(int i=0; i<lists.size(); i++){ if(lists[i]!=NULL){ if(mini < 0){ mini = i; }else if(lists[i]->val<lists[mini]->val){ mini = i; } } } if(mini<0) break; cur->next = lists[mini]; cur = cur->next; lists[mini] = lists[mini]->next;//注意要修改lists数组中的值,不能直接提取指针然后修改 } return dummy.next; }};
Solution(2): 两个两个的merge,直到将所有的链表都merge到一起,时间复杂度大约是O((nk)/2),Leetcode运行时间222ms,因为leetcode给的时间限制不严格,这样做也可以过。复用Merge Two Sorted Lists的函数。
Code:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* mergeKLists(vector<ListNode*>& lists) { ListNode* start = NULL; for(int i=0; i<lists.size(); i++){ start = mergeTwoLists(start, lists[i]); } return start; }private: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if(l1==NULL) return l2; else if(l2==NULL) return l1; ListNode* start = NULL; if(l1->val < l2->val){ start = l1; l1 = l1->next; }else{ start = l2; l2 = l2->next; } ListNode* cur = start; while(l1!=NULL && l2!=NULL){ if(l1->val < l2->val){ cur->next = l1; l1 = l1->next; }else{ cur->next = l2; l2 = l2->next; } cur = cur->next; } if(l1!=NULL) cur->next = l1; else if(l2!=NULL) cur->next = l2; return start; }};
Solution(3): 将上面的做法加以改进,使用分治法,可以得到O(nlogn)的解法,Leetcode运行时间26ms。
Code:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* mergeKLists(vector<ListNode*>& lists) { return mergeKLists(lists.begin(), lists.end()); }private: ListNode* mergeKLists(vector<ListNode*>::iterator itbegin, vector<ListNode*>::iterator itend){ if(itbegin>=itend) return NULL; if((itend-itbegin)==1) return *itbegin; if((itend-itbegin)==2) return mergeTwoLists(*itbegin, *(itbegin+1)); auto mid = itbegin + (itend-itbegin)/2; return mergeTwoLists(mergeKLists(itbegin, mid),mergeKLists(mid, itend)); } ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {}//同上,我就不写了};
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