LeetCode :: Convert Sorted Array (link list) to Binary Search Tree [tree]

1.Given an array where elements are sorted in ascending order, convert it to a height balanced BST.





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





这里两道题目。是连在一起的两题。给你一个排好序（升序）的数组或者链表，将它们转为一棵平衡二叉树。如果不排好序的话。一组随机输入的数据，就必须採用RBT或者AVL树，这样操作会变得更复杂。涉及到旋转，可是这里排好序了。所以，仅仅要找到中位数。作为根，然后递归地依据中位数的左、右数列来构建左右子树；





两题的思路都是如上所述， 唯一的差别就是，链表寻找中位数会麻烦一些。须要引入fast、slow两个指针，来寻找中位数。代码例如以下：

1.Array

class Solution {
public:
    TreeNode *Tree(int left, int right, vector<int> &num){
        TreeNode *root = NULL;
        if (left <= right){
            int cen = (left + right) / 2;
            root = new TreeNode(num[cen]);
            root->left = Tree(left, cen - 1, num);
            root->right = Tree(cen + 1, right, num);
        }
        return root;
    }

    TreeNode *sortedArrayToBST(vector<int> &num) {
        TreeNode *T = NULL;
        int N = num.size();
        T = Tree(0, N - 1, num);
        return T;
    }
};

2.link-list

class Solution {
public:
    ListNode *findMid(ListNode *head){        //这里假设链表中仅仅有两个数字。则mid返回的是head->next.
        if (head == NULL || head -> next == NULL)
            return head;
        ListNode *fast, *slow, *pre;
        fast = slow = head;
        pre = NULL;
        while (fast && fast->next){
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = NULL;
        return slow;
    }

    TreeNode *buildTree(ListNode *head){
        TreeNode *root = NULL;
        ListNode *mid = NULL;
        if (head){
            mid = findMid(head);
            root = new TreeNode(mid->val);
            if (head != mid){
                root->left = buildTree(head);
                root->right = buildTree(mid->next);
            }
        }
        return root;

    }

    TreeNode *sortedListToBST(ListNode *head) {
        TreeNode *T;
        T = buildTree(head);
        return T;
    }
};