【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);
    }
};