Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
這題我一看到，覺得跟上題(108)一樣啊，就是找中間結(jié)點(diǎn)遞歸生成BST。于是我先用了個(gè)笨方法先用O(n)時(shí)間把list上的元素?cái)]下來保存到nums里，跑了一下，AC了。這方法可行，但占用了O(n)空間存儲(chǔ)。
不好的方法
//不好的方法
public TreeNode sortedListToBST(ListNode head) {
if (head == null) return null;
List<Integer> list = new ArrayList<>();
while (head != null) {
list.add(head.val);
head = head.next;
}
int[] nums = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
nums[i] = list.get(i);
}
return recursion(nums, 0, nums.length - 1);
}
private TreeNode recursion(int nums[], int left, int right) {
if (left > right) return null;
int mid = (left + right) / 2;
TreeNode root = new TreeNode(nums[mid]);
root.left = recursion(nums, left, mid - 1);
root.right = recursion(nums, mid + 1, right);
return root;
}
好的方法：
還是看到SOLUTIONS里的高票答案，直接構(gòu)造BST，中點(diǎn)用快慢指針尋找。
注意：
if (head == tail) return null;這個(gè)判斷條件不能少。while里的判斷條件可以看成一個(gè)routine了，而且是!=tail，不是null
public TreeNode sortedListToBST(ListNode head) {
if (head == null) return null;
return recursion(head, null);
}
private TreeNode recursion(ListNode head, ListNode tail) {
ListNode runner = head;
ListNode walker = head;
//這個(gè)return不能少
if (head == tail) return null;
//這while里的判斷條件可以看成一個(gè)routine了，而且是!=tail，不是null
while (runner.next != tail && runner.next.next != tail) {
walker = walker.next;
runner = runner.next.next;
}
TreeNode root = new TreeNode(walker.val);
root.left = recursion(head, walker);
root.right = recursion(walker.next, tail);
return root;
}
