跳到主要内容

Convert Sorted List to Binary Search Tree

描述

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

分析

这题与上一题类似,但是单链表不能随机访问,而自顶向下的二分法必须需要 RandomAccessIterator,因此前面的方法不适用本题。

存在一种自底向上(bottom-up)的方法,见 http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html

分治法,自顶向下

分治法,类似于 Convert Sorted Array to Binary Search Tree,自顶向下,复杂度 O(nlogn)

// Convert Sorted List to Binary Search Tree
// 二分法,类似于 Convert Sorted Array to Binary Search Tree,
// 自顶向下,时间复杂度O(nlogn),空间复杂度O(logn)
class Solution {
public TreeNode sortedListToBST (ListNode head) {
if(head == null) return null;
if(head.next == null) return new TreeNode(head.val);

ListNode mid = cutAtMiddle(head);

TreeNode root = new TreeNode(mid.val);
root.left = sortedListToBST(head);
root.right = sortedListToBST(mid.next);

return root;
}

ListNode cutAtMiddle(ListNode head) {
if(head == null) return null;

ListNode fast = head;
ListNode slow = head;
ListNode prev_slow = head;

while(fast != null && fast.next != null){
prev_slow = slow;
slow = slow.next;
fast = fast.next.next;
}

prev_slow.next = null;
return slow;
}
}

自底向上

// Convert Sorted List to Binary Search Tree
// bottom-up,时间复杂度O(n),空间复杂度O(logn)
public class Solution {
public TreeNode sortedListToBST(ListNode head) {
int len = 0;
ListNode p = head;
while (p != null) {
len++;
p = p.next;
}
return sortedListToBST(new Container(head), 0, len - 1);
}
private static TreeNode sortedListToBST(Container list, int start, int end) {
if (start > end) return null;

int mid = start + (end - start) / 2;
TreeNode leftChild = sortedListToBST(list, start, mid - 1);
TreeNode parent = new TreeNode(list.p.val);
parent.left = leftChild;
list.p = list.p.next;
parent.right = sortedListToBST(list, mid + 1, end);
return parent;
}
// 模拟二级指针
static class Container {
ListNode p;
public Container(ListNode p) {
this.p = p;
}
}
}

相关题目