LeetCode | Convert Sorted List to Binary Search Tree

题目

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

分析

递归不断构造左右子树

普通构造方法（解法1）就是每次递归调用都要去遍历找到中间的节点。

高端点的方法（解法2），类似中序遍历树的写法，先构造左子树，左子树构造完时，链表上的指针刚好指向中间节点，再构造右子树即可。

解法1

	public TreeNode sortedListToBST(ListNode head) {
		if (head == null) {
			return null;
		}

		int n = 0;
		ListNode p = head;
		while (p != null) {
			++n;
			p = p.next;
		}
		return solve(head, n - 1);
	}

	private TreeNode solve(ListNode head, int high) {
		if (high < 0) {
			return null;
		}
		int mid = high / 2;
		ListNode p = head;
		for (int i = 0; i < mid; ++i) {
			p = p.next;
		}
		TreeNode root = new TreeNode(p.val);
		root.left = solve(head, mid - 1);
		root.right = solve(p.next, high - mid - 1);
		return root;
	}

解法2

public class ConvertSortedListToBinarySearchTree {
	private ListNode head;

	public TreeNode sortedListToBST(ListNode head) {
		if (head == null) {
			return null;
		}

		this.head = head;
		int n = 0;
		ListNode p = head;
		while (p != null) {
			++n;
			p = p.next;
		}
		return solve(0, n - 1);
	}

	private TreeNode solve(int low, int high) {
		if (low > high) {
			return null;
		}
		int mid = low + (high - low) / 2;
		TreeNode left = solve(low, mid - 1);
		TreeNode root = new TreeNode(head.val);
		head = head.next;
		TreeNode right = solve(mid + 1, high);
		root.left = left;
		root.right = right;
		return root;
	}
}