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109. Convert Sorted List to Binary Search Tree

MediumOpen on LeetCodeProblem statement

Problem Statement

109. Convert Sorted List to Binary Search Tree

Medium

Given the head of a singly linked list where elements are sorted in ascending order, convert it to a height-balanced binary search tree.

 

Example 1:

Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.

Example 2:

Input: head = []
Output: []

 

Constraints:

C++

Source file
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head)
    {
    	return sortedListToBST( head, NULL );
    }
    
private:
    TreeNode *sortedListToBST(ListNode *head, ListNode *tail)
    {
    	if( head == tail )
    		return NULL;
    	if( head->next == tail )    // 
    	{	
    		TreeNode *root = new TreeNode( head->val );
    		return root;
    	}
    	ListNode *mid = head, *temp = head;
    	while( temp != tail && temp->next != tail )    // 寻找中间节点
    	{
    		mid = mid->next;
    		temp = temp->next->next;
    	}
    	TreeNode *root = new TreeNode( mid->val );
    	root->left = sortedListToBST( head, mid );
    	root->right = sortedListToBST( mid->next, tail );
    	return root;
    }
};