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);}
};