代码随想录算法训练营第二十天-二叉树-108.将有序数组转换为二叉搜索树

• 构造的关键点就是找到树的根结点
• 每个子树也是找其的根结点
• 注意如果构造一个线性二叉树也是搜索树，但不符合题意

#include <iostream>
#include <vector>

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(): val(0), left(nullptr), right(nullptr) {}
    TreeNode(int v): val(v), left(nullptr), right(nullptr) {}
    TreeNode(int v, TreeNode* l, TreeNode* r): val(v), left(l), right(r) {}
};

class Solution {
private:
    TreeNode* traversal(std::vector<int>& nums, int left, int right) {
        if (left > right)
            return nullptr;
        int mid = (left + right) / 2;
        TreeNode* root = new TreeNode(nums.at(mid));
        root->left = traversal(nums, left, mid - 1);
        root->right = traversal(nums, mid + 1, right);
        return root;
    }
public:
    TreeNode* sortedArrayToBST(std::vector<int>& nums) {
        return traversal(nums, 0, nums.size() - 1);
    }
};
int main()
{
    Solution s;
    return 0;
}

• 汇总