Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
题意
有序数组转化为二叉搜索树
思路1
- 始终选择中间位置左边元素作为根节点然后分别构造左右子树
代码1
/**
* 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 {
private:
vector<int> nums;
public:
TreeNode* sortedArrayToBST(vector<int>& nums) {
this->nums = nums;
return helper(0, nums.size() - 1);
}
TreeNode* helper(int left, int right){
if(left > right) return NULL;
// 始终选择中间位置左边元素作为根节点
int p = (left + right) / 2;
TreeNode *root = new TreeNode(nums[p]);
root->left = helper(left, p - 1);
root->right = helper(p + 1, right);
return root;
}
};