LeetCode题解——Convert Sorted List to Binary Search Tree

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

分析:要求最后得到一颗平衡二叉树，那么链表的中间节点就是二叉树的根节点，中间节点左边就是二叉树的左子树，中间节点右边就是二叉树的右子树

首先找到链表的中间节点作为BST的根节点，根节点的左子树是以中间节点左边的节点建立的一颗平衡二叉树，根节点的右子树是以中间节点右边的节点建立的一颗平衡二叉树，因此递归的解决此问题。

/**
 * 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) {//O(n^2)解法
        if(!head) return NULL;
        //find the median of the listNode O(n)
        ListNode*  MidNode = findMedianOfList(head);
        TreeNode* root=new TreeNode(MidNode->val);
        root->right = sortedListToBST(MidNode->next);
	// set MidNode to NULL get the left part of the List
	ListNode* phead = new ListNode(0);
	phead->next = head;
	ListNode* newhead = phead;
	while(newhead->next && newhead->next!=MidNode){
		newhead = newhead->next;
	}
	newhead->next = NULL;
        root->left = sortedListToBST(phead->next);
        return root;
    }
    ListNode* findMedianOfList(ListNode* head){
        if(!head||!head->next) return head;
        ListNode* fast = head, *slow = head;
        while(fast && fast->next){
            fast = fast->next->next;
            slow = slow->next;
        }
        return slow;
    }
};