Convert Sorted List to Binary Search Tree

最新推荐文章于 2020-08-18 08:24:28 发布

原创最新推荐文章于 2020-08-18 08:24:28 发布 · 725 阅读

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本文详细介绍了如何通过中序遍历来将一个升序排序的单链表转换为一个高度平衡的二叉搜索树，并提供了关键步骤和代码实现。

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

解题思路:采用中序排列的方法递归地决定每个结点的数值；

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
//Definition for singly - linked list.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
	
};
//Definition for binary tree
struct TreeNode {
   int val;
   TreeNode *left;
   TreeNode *right;
   TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
void ListToBST(vector<ListNode*>::iterator Begin, vector<ListNode*>::iterator End, TreeNode*&root)
{
	if (Begin==End)
		return;
	auto len = distance(Begin,End);
	if ((int)len == 1){
		root = new TreeNode((*Begin)->val);
		return;
	}
	auto Iter_mid = Begin;
	advance(Iter_mid, len / 2);
	root = new TreeNode((*Iter_mid)->val);
	ListToBST(Begin, Iter_mid, root->left);
	ListToBST(Iter_mid + 1, End, root->right);
}
TreeNode *sortedListToBST(ListNode *head) {
	vector<ListNode*>NodeVector;
	for (; head != NULL; head = head->next)
		NodeVector.push_back(head);
	TreeNode* Root = NULL;
	ListToBST(NodeVector.begin(), NodeVector.end(), Root);
	return Root;
}