【力扣】108.将有序数组转换为二叉搜索树

AC截图

题目

思路

因为nums数组是严格递增的，所以只需要每次选出中间节点，然后用左边部分构建左子树，用右边部分构建右子树。

代码

/*** 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* sortToBST(vector<int>& nums,int left,int right){if(left>right){return NULL;}int mid = (left+right)/2;TreeNode* root = new TreeNode(nums[mid]);root->left = sortToBST(nums,left,mid-1);root->right = sortToBST(nums,mid+1,right);return root;}TreeNode* sortedArrayToBST(vector<int>& nums) {return sortToBST(nums,0,nums.size()-1);}
};