题意:将有序数列转化成二叉搜索树。
思路:DFS,注意中间结点的处理。
/**
* 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:
vector<int> n;
TreeNode* sortedArrayToBST(vector<int>& nums) {
n = nums;
if(nums.size() == 0) return NULL;
return dfs(0, nums.size() - 1);
}
TreeNode* dfs(int l, int r) {
if(l > r) return NULL;
int mid = l + (r - l) / 2;
TreeNode* root = new TreeNode(n[mid]);
root->left = dfs(l, mid - 1);
root->right = dfs(mid + 1, r);
return root;
}
};