(M)DFS:109. Convert Sorted List to Binary Search Tree
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这道题的意思是,给一个排序过的单链表,把这个单链表转变成一棵二叉搜索树。
这道题的关键在于,怎样找到单链表的中间那个节点。经典的方法是使用快慢指针,也就是每次让快指针走两步,让慢指针走一步,这样当快指针走到末尾的时候,慢指针指的就是链表的中间节点。然后把链表分成了两部分,左右两部分分别dfs
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; *//** * 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: TreeNode* sortedListToBST(ListNode* head) { if(head == NULL) return NULL; return dfs(head, NULL); } TreeNode* dfs(ListNode* start, ListNode* end) //不包括end节点的 { ListNode* fast = start, *slow = start; if(start == end) return NULL; while(fast != end && fast->next != end) { fast = fast->next->next; slow = slow->next; } TreeNode* newtree = new TreeNode(slow->val); newtree->left = dfs(start, slow); newtree->right = dfs(slow->next, end); return newtree; }};
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