面试题13：有序单链表或者数组转平衡二叉搜索树

• Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
• Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * 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 *sortedListToBST(ListNode *head) {
        if(head == nullptr) return nullptr;

        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* slow_pre = nullptr;
        while(fast && fast->next){
            slow_pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        if(slow_pre){
            slow_pre->next = nullptr;
        }
        
        TreeNode* root = new TreeNode(slow->val);
        if(slow!= head)
            root->left = sortedListToBST(head);
        else
            root->left = nullptr;
        root->right = sortedListToBST(slow->next);

        return root;
    }
};

/**
 * 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) {
        int n = num.size();
        return sortedArrayToBST(num,0,n-1);
    }

    TreeNode* sortedArrayToBST(vector<int>& num,int l,int r){
        if(l > r) return nullptr;
        int mid = l + (r - l+1)/2;  //加1保证左子树节点个数不少于右边的节点个数，不加1也是平衡二叉树搜索树，但是无法通过验证
        TreeNode* root = new TreeNode(num[mid]);
        root->left = sortedArrayToBST(num,l,mid - 1);
        root->right = sortedArrayToBST(num,mid+1,r);
        return root;
    }

};