【LeetCode108】 Convert Sorted Array to Binary Search Tree

20191209更新：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        if(nums.empty())
            return NULL;
        else if(nums.size() == 1)
            return new TreeNode(nums[0]);
        else{
            int mid = nums.size()/2;
            TreeNode* node = new TreeNode(nums[mid]);
            // node->val = nums[mid]; 
            vector<int> left_vector(nums.begin(), nums.begin()+mid);
            vector<int> right_vector(vector<int>(nums.begin()+mid+1, nums.end()));
            node->left = sortedArrayToBST(left_vector);
            node->right = sortedArrayToBST(right_vector);
            return node;
        }
    }
};
           

nums.size()==1必须要单独讨，不然用(nums.begin()， nums.begin()+mid)初始化会出错；

指针也要特别注意，用new来调用初始化函数最好，指针指向的东西进行了初始化。

时间和内存开销都很大，而且是参考了上次的才做出来，需要好好打磨。

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        if(nums.size() == 0)//有可能直接输入空数组
            return NULL;
        if(nums.size() == 1)//边界条件
            return new TreeNode(nums[0]);
        int inter = nums.size()/2;
        TreeNode* node = new TreeNode(nums[inter]);//使用new语句返回该节点的地址，而不是直接创建
        vector<int> left_array(nums.begin(), nums.begin()+inter);//创建子数组
        vector<int> right_array(nums.begin()+inter+1, nums.end());
        node->left = sortedArrayToBST(left_array);//node是指针而不是节点本身，所以要用->
        node->right = sortedArrayToBST(right_array);
        return node;
    }
};