今日算法之_137_有序链表转化为二叉树
前言
Github:https://github.com/HealerJean
1、有序链表转化为二叉树
给定一个单链表,其中的元素按升序排序,将其转换为高度平衡的二叉搜索树。
本题中,一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。
示例:
给定的有序链表: [-10, -3, 0, 5, 9],
一个可能的答案是:[0, -3, 9, -10, null, 5], 它可以表示下面这个高度平衡二叉搜索树:
0
/ \
-3 9
/ /
-10 5
1.1、解题思路
先转换为有序数组,再转化
1.2、算法
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 = list.stream().mapToInt(Integer::intValue).toArray();
return inSort(nums, 0, nums.length-1);
}
private TreeNode inSort(int[] nums, int start, int end) {
if (start > end) {
return null;
}
int index = (start + end) / 2;
TreeNode treeNode = new TreeNode(nums[index]);
treeNode.left = inSort(nums, start, index - 1);
treeNode.right = inSort(nums, index + 1, end);
return treeNode;
}
1.3、测试
@Test
public void test() {
sortedListToBST(listNode());
}
public ListNode listNode() {
ListNode listNode_5 = new ListNode(9, null);
ListNode listNode_4 = new ListNode(5, listNode_5);
ListNode listNode_3 = new ListNode(0, listNode_4);
ListNode listNode_2 = new ListNode(-3, listNode_3);
ListNode listNode_1 = new ListNode(-10, listNode_2);
return listNode_1;
}
public String listNodeStr(ListNode listNode, String str) {
if (listNode == null) {
return str.substring(0, str.lastIndexOf(","));
}
str = str + listNode.val + ",";
return listNodeStr(listNode.next, str);
}
class ListNode {
int val;
ListNode next;
public ListNode(int val) {
this.val = val;
}
public ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}
public TreeNode initTreeNode() {
TreeNode treeNode1 = new TreeNode(3, null, null);
TreeNode treeNode2 = new TreeNode(6, null, null);
TreeNode treeNode3 = new TreeNode(4, treeNode1, treeNode2);
TreeNode treeNode4 = new TreeNode(1, null, null);
TreeNode root = new TreeNode(5, treeNode3, treeNode4);
return root;
}
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
TreeNode(int x, TreeNode left, TreeNode right) {
this.val = x;
this.left = left;
this.right = right;
}
}