109. Convert Sorted List to Binary Search Tree
摘要：
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-bal...
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted linked list: [-10,-3,0,5,9], One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST: 0 / \ -39 // -105
難度：medium
題目：給定一個單鏈表其元素為升序排列，將其轉換成高度平衡的二叉搜尋樹
思路：中序遍歷
Runtime: 1 ms, faster than 99.17% of Java online submissions for Convert Sorted List to Binary Search Tree.
Memory Usage: 41 MB, less than 9.56% of Java online submissions for Convert Sorted List to Binary Search Tree.
/**
* Definition for singly-linked list.
* public class ListNode {
*int val;
*ListNode next;
*ListNode(int x) { val = x; }
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
*int val;
*TreeNode left;
*TreeNode right;
*TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode sortedListToBST(ListNode head) {
if (null == head) {
return null;
}
ListNode ptr = head;
int count = 0;
for (; ptr != null; ptr = ptr.next, count++);
ListNode[] headList = {head};
return sortedListToBST(headList, 0, count - 1);
}
public TreeNode sortedListToBST(ListNode[] head, int start, int end) {
if (start > end) {
return null;
}
int mid = start + (end - start) / 2;
TreeNode left = sortedListToBST(head, start, mid - 1);
TreeNode root = new TreeNode(head[0].val);
root.left = left;
head[0] = head[0].next;
root.right = sortedListToBST(head, mid + 1, end);
return root;
}
}