[LeetCode] Convert Sorted List to Binary Search Tree(分治)

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

方法:为了使BST高度平衡,要找链表中的中值作为当前根节点。

/**
 * 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) {
        if(head==NULL)
            return NULL;
        ListNode *pmid = FindMid(head);
        TreeNode *root = new TreeNode(pmid->val);
        if(pmid != head)
           root->left = sortedListToBST(head);
        root->right = sortedListToBST(pmid->next);
        return root;
    }
private:
    ListNode *FindMid(ListNode *head){//找中间结点
        if(head->next==NULL)
            return head;
        ListNode *Pmid=head,*p=head,*Pmin_pre = head;
        while(p!=NULL && p->next != NULL)
        {
           p = p->next;
           p = p->next;
           Pmin_pre = Pmid;
           Pmid = Pmid->next;
        }//end while;
        Pmin_pre->next = NULL;
        return Pmid; 
    }
};
原文地址:https://www.cnblogs.com/Xylophone/p/3890549.html