LeetCode: Convert Sorted List to Binary Search Tree

第一次想用懒汉方法把list弄成vector再照前一题的方法，结果被判Memory exceed了。。然后只好再写次

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *dfs(ListNode* head, int len) {
         if (!len) return NULL;
         int tmplen = len/2;
         ListNode *tmpl = head;
         while (tmplen && tmpl) {
             tmpl = tmpl->next;
             tmplen--;
         }
         TreeNode *tmp = new TreeNode(tmpl->val);
         tmp->left = dfs(head, len/2);
         tmp->right = dfs(tmpl->next, (len-1)/2);
         return tmp;
    }
    TreeNode *sortedListToBST(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!head) return NULL;
        int len = 0;
        ListNode *tmp = head;
        while (tmp) {
            len++;
            tmp = tmp->next;
        }
        return dfs(head, len);
    }
};

 后来写了一个用双指针的写法。Interview的话应该这个程序更好

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if (head == NULL) return NULL;
        ListNode *mid = head;
        ListNode *end = head;
        ListNode *pre = new ListNode(0);
        pre->next = head;
        while (end) {
            end = end->next;
            if (end == NULL) break;
            end = end->next;
            mid = mid->next;
            pre = pre->next;
        }
        TreeNode *res = new TreeNode(mid->val);
        end = mid->next;
        if (pre->next == head) return new TreeNode(head->val);
        else pre->next = NULL;
        res->left = sortedListToBST(head);
        res->right = sortedListToBST(end);
        return res;
    }
};

 C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode SortedListToBST(ListNode head) {
        if (head == null) return null;
        ListNode mid = head;
        ListNode end = head;
        ListNode pre = new ListNode(0);
        pre.next = head;
        while (end != null) {
            end = end.next;
            if (end == null) break;
            end = end.next;
            mid = mid.next;
            pre = pre.next;
        }
        TreeNode ans = new TreeNode(mid.val);
        end = mid.next;
        if (pre.next == head) return new TreeNode(head.val);
        else pre.next = null;
        ans.left = SortedListToBST(head);
        ans.right = SortedListToBST(end);
        return ans;
    }
}