109. Convert Sorted List to Binary Search Tree

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每日一题 2019 - 04 - 25

题目：

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

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Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

解法：

这个题让在给定的链表序列里面构造出一颗平衡二叉搜索树，也即每一颗子树的左右子树高度差不大于 1 ，由于给定的序列是排序好的序列，所以可以直接使用二分法来完成构建；

  • 每一次取出当前序列的中点，把当前序列的中点当做根节点，然后使用左右两边的序列对左右两边的子树进行构建；
  • 重复上述过程，直到左右两边的序列遍历完毕，即可完成任务；

代码：

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/**
 * 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) {
        if(head == NULL)
        {
            return NULL;
        }
        vector<int> test ;
        ListNode* temp = head;
        while(temp)
        {
            test.push_back(temp->val);
            temp = temp -> next ; 
        }
        int mid = (0+test.size())/2 ;
        TreeNode* node = new TreeNode(test[mid]);
        node -> left = build(test,0,mid-1);
        node -> right = build(test,mid+1,test.size()-1);
        return node;
    }
    TreeNode* build(vector<int>& test,int start,int end)
    {
        if(start > end || start < 0 || end < 0 )
        {
            return NULL ;
        }
        if(start == end)
        {
            TreeNode* now = new TreeNode(test[start]);
            return now ;
        }
        int mid = (start + end) /2 ;
        TreeNode* now = new TreeNode(test[mid]);
        now -> left = build(test,start,mid-1);
        now -> right = build(test,mid+1,end);
        return now ;
    }
};
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