LintCode-排序列表转换为二分查找树

标签：lintcode   面试   

分析：就是一个简单的递归，只是需要有些链表的操作而已

代码：

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: a tree node
     */
    TreeNode *sortedListToBST(ListNode *head) {
        // write your code here
        if(head==nullptr)
            return nullptr;
        int len = 0;
        ListNode*temp = head;
        while(temp){len++;temp = temp->next;};
        if(len==1)
        {
            return new TreeNode(head->val);
        }
        else if(len==2)
        {
            TreeNode*root = new TreeNode(head->val);
            root->right = new TreeNode(head->next->val);
            return root;
        }
        else
        {
            len/=2;
            temp = head;
            int cnt = 0;
            while(cnt<len)
            {
                temp = temp->next;
                cnt++;
            }
            ListNode*pre = head;
            while(pre->next!=temp)
                pre = pre->next;
            pre->next = nullptr;
            TreeNode*root = new TreeNode(temp->val);
            root->left = sortedListToBST(head);
            root->right = sortedListToBST(temp->next);
            return root;
            
        }
    }
};

LintCode-排序列表转换为二分查找树

标签：lintcode   面试   

原文地址：http://blog.csdn.net/wangyuquanliuli/article/details/47359405