lintcode-106-排序列表转换为二分查找树

106-排序列表转换为二分查找树

给出一个所有元素以升序排序的单链表，将它转换成一棵高度平衡的二分查找树

样例

标签

递归 链表

思路

类似于二分查找，每次将链表二分，中间节点作为根节点，在建立左子树与右子树，递归即可

code

/**
 * 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
        TreeNode *root = NULL;

        if(head != NULL) {
            ListNode *left = NULL, *right = NULL;
            root = new TreeNode(dichotomyList(head, left, right));
            root->left = sortedListToBST(left);
            root->right = sortedListToBST(right);
        }

        return root;
    }

    int dichotomyList(ListNode *head, ListNode *&left, ListNode *&right) {
        if(head->next != NULL) {
            ListNode *fast = head, *slow = head, *temp = head;
            while(fast != NULL && fast->next != NULL) {
                temp = slow;
                slow = slow->next;
                fast = fast->next->next;
            }
            temp->next = NULL;

            left = head;
            right = slow->next;

            return slow->val;
        }
        else {
            left = NULL;
            right = NULL;
            return head->val;
        }
    }
};