static void Main(string[] args)
{
ListNode l1 = new ListNode(1);
l1.next = new ListNode(2);
l1.next.next = new ListNode(3);
l1.next.next.next = new ListNode(4);
l1.next.next.next.next = new ListNode(5);
l1.next.next.next.next.next = new ListNode(6);
l1.next.next.next.next.next.next = new ListNode(7);
TreeNode head = sortedListToBST(l1);
}
private static TreeNode sortedListToBST(ref ListNode list, int start, int end)
{
if (start > end)
return null;
int mid = start + (end - start) / 2;
TreeNode leftChild = sortedListToBST(ref list, start, mid - 1);
TreeNode parent = new TreeNode(list.val);
parent.left = leftChild;
list = list.next;
parent.right = sortedListToBST(ref list, mid + 1, end);
return parent;
}
/*
* reference:
* https://github.com/soulmachine/leetcode
*
* analysis from the above blog:
* LeetCode, Convert Sorted List to Binary Search Tree
bottom-up，时间复杂度O(n)，空间复杂度O(logn)
*
* understand the analysis from the blog:
* http://codeganker.blogspot.ca/2014/11/leetcode.html
*
* 先来看看最简单的Convert Sorted Array to Binary Search Tree，数组本身是有序的，那么我们知道每次只要取中点作为根，
* 然后递归构建对应的左右子树就可以了，递归的写法跟常规稍有不同，就是要把根root先new出来，然后它的左节点接到递归左边
* 部分的返回值，右节点接到递归右边部分的返回值，最后将root返回回去。这个模板在树的构造中非常有用，其他几道题也都是按照这个来实现。
接下来是Convert Sorted List to Binary Search Tree，这个跟Convert Sorted Array to Binary Search Tree比较近似，
* 区别是元素存储的数据结构换成了链表，不过引入了一个重要的问题，就是链表的访问不是随机存取的，也就是不是O(1)的，如果每次
* 去获取中点，然后进行左右递归的话，我们知道得到中点是O(n/2)=O(n)的，如此递推式是T(n) = 2T(n/2)+n/2，复杂度是O(nlogn)，
* 并不是线性的，所以这里我们就得利用到树的中序遍历了，按照递归中序遍历的顺序对链表结点一个个进行访问，而我们要构造的二分查找树
* 正是按照链表的顺序来的。如此就能按照链表的访问顺序来构造，不会因此而增加找中间结点的复杂度。
*
* julia's comment: (08/25/2015)
* 1. Try to use ref in the function argument of list
* 2. Read the website:
* http://stackoverflow.com/questions/4311226/list-passed-by-ref-help-me-explain-this-behaviour
* 3. Online judge:
* 32 / 32 test cases passed.
Status: Accepted
Runtime: 176 ms
*
* 9/21/2015
* 1. Totally suprised, cannot recall this bottom up technical solution by just reading less then 30 lines of code.
* 2. So, Julia reviewed the article and got the idea back to her head. In other words, last time, she learned,
* then, she could not memorize the analysis, the main idea. Bottom up, the idea to get the root node using
* O(1) time instead of naive solution O(n) to go through the list node one by one.
* 3. Here is the article she read:
* http://articles.leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html
* 4. Here are main ideas from the above blog:
* Best Solution:
As usual, the best solution requires you to think from another perspective. In other words, we no longer
* create nodes in the tree using the top-down approach. We create nodes bottom-up, and assign them to its parents.
* The bottom-up approach enables us to access the list in its order while creating nodes.
Isn’t the bottom-up approach neat? Each time you are stucked with the top-down approach, give bottom-up a try.
* Although bottom-up approach is not the most natural way we think, it is extremely helpful in some cases. However,
* you should prefer top-down instead of bottom-up in general, since the latter is more difficult to verify in correctness.
Below is the code for converting a singly linked list to a balanced BST. Please note that the algorithm requires
* the list’s length to be passed in as the function’s parameters. The list’s length could be found in O(N) time
* by traversing the entire list’s once. The recursive calls traverse the list and create tree’s nodes by the list’s
* order, which also takes O(N) time. Therefore, the overall run time complexity is still O(N).
*
* Here is the C++ code for Julia to review the solution and the idea:
BinaryTree* sortedListToBST(ListNode *& list, int start, int end) {
if (start > end) return NULL;
// same as (start+end)/2, avoids overflow
int mid = start + (end - start) / 2;
BinaryTree *leftChild = sortedListToBST(list, start, mid-1);
BinaryTree *parent = new BinaryTree(list->data);
parent->left = leftChild;
list = list->next;
parent->right = sortedListToBST(list, mid+1, end);
return parent;
}
BinaryTree* sortedListToBST(ListNode *head, int n) {
return sortedListToBST(head, 0, n-1);
}
5. Try to memorize the solution, it is the best strategy. Need to work out more on different ideas to solve the tree problems.
* Top down, bottom up, what is difference?
*
* Learn a lesson today, 9/21/2015, after 1 month's practice.
*
* Need to review other solutions for question 109.
*
*/
public static TreeNode sortedListToBST(ListNode head)
{
int len = 0;
ListNode p = head;
while (p != null)
{
len++;
p = p.next;
}
return sortedListToBST(ref head, 0, len - 1);
}