Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *sortedArrayToBST(vector<int> &num) {
if(num.empty())
return NULL;
int mid=(num.size()-1)/2;
TreeNode *root=new TreeNode(num[mid]);
vector<int> leftArray(num.begin(),num.begin()+mid);
vector<int> rightArray(num.begin()+mid+1,num.end());
if(mid>0)
root->left=sortedArrayToBST(leftArray);
root->right=sortedArrayToBST(rightArray);
return root;
}
};
本文介绍了一种将已排序的数组转换为高度平衡的二叉搜索树的方法。通过递归方式选取数组中间元素作为根节点,并将左右两侧的子数组分别构造为左子树和右子树。
339

被折叠的 条评论
为什么被折叠?



