因为是有序数组,取中间位置的值为树根。将数组分成两半。 对左半递归构建左子树。对右半递归构建右子树。
就是一个二分法。
class Solution {
public:
TreeNode* sortedArrayToBST(vector<int>& nums) {
return ToBST(nums, 0, nums.size()-1);
}
TreeNode* ToBST(vector<int>& nums, int left, int right){
if(left > right) return NULL;
int mid = (int)(left + right + 1) / 2; (注意这里要+1 ,因为int 将小数位给舍弃掉了),
TreeNode* root = new TreeNode(nums[mid]);
cout << "nums[mid] is:" << nums[mid] << "mid is:" << mid << endl;
root->left = ToBST(nums, left, mid-1);
root->right = ToBST(nums, mid+1, right);
return root;
}
};