LeetCode -- Convert SortedList To BST
题目描述：
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
就是把链表转化为二叉查找树
思路：
使用分治策略
1.把链表节点遍历，存在nodes集合中
2.用[0, length/2)节点创建左子树，用(length/2,n]节点创建右子树，使用nodes[length/2]来创建当前节点。
实现代码：
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
就是把链表转化为二叉查找树
思路：
使用分治策略
1.把链表节点遍历，存在nodes集合中
2.用[0, length/2)节点创建左子树，用(length/2,n]节点创建右子树，使用nodes[length/2]来创建当前节点。
实现代码：
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode SortedListToBST(ListNode head) {
if(head == null){
return null;
}
var nodes = new List<int>();
while(head != null){
nodes.Add(head.val);
head = head.next;
}
TreeNode root = null;
BuildTree(nodes,ref root);
return root;
}
private void BuildTree(IList<int> values,ref TreeNode n){
if(values.Count == 0){
return;
}
var mid = (int)(values.Count/2);
var self = values[mid];
n = new TreeNode(self);
var leftVals= values.Where(x=>x < self);
var rightVals = values.Where(x=>x > self);
if(leftVals.Any()){
BuildTree(leftVals.ToList() ,ref n.left);
}
if(rightVals.Any()){
BuildTree(rightVals.ToList(),ref n.right);
}
}
}