leetcode 109. 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.

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:

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

思路:类似线段树，每次去区间中间那个值作为跟节点，注意出现2个数的时候我们先取右边那个，也就是比较大的那个数。

这段代码飞快。

/**
 * 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:
    vector<int> v;
    TreeNode* dfs(int l, int r) {
        if (l > r) return nullptr;
        int mid = ((l + r)%2 == 0) ? (l+r)/2 : (l+r)/2+1;
        TreeNode* root = new TreeNode(v[mid]);
        //root->val = ;
        root->left = dfs(l, mid-1);
        root->right = dfs(mid+1, r);
        return root;
    }
    TreeNode* sortedListToBST(ListNode* head) {
        ListNode* p;
        p = head;
        while (p) {
            v.push_back(p->val);
            p = p->next;
        }
        int l = 0, r = v.size()-1;
        return dfs(l, r);
    }
};