leetcode108. 将有序数组转换为二叉搜索树

分析

为了保持平衡，根节点每次都取中点。然后将剩余的区间分别给左右子树分派。

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

};