Convert Sorted List to Binary Search Tree

最新推荐文章于 2020-08-18 08:24:28 发布

zzyazzy

最新推荐文章于 2020-08-18 08:24:28 发布

阅读量503

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/zzyafyj/article/details/41805553

本文介绍了一种将升序单链表转换为高度平衡二叉搜索树的方法。通过计算链表长度并找到中间节点，递归地构建左子树和右子树，最终形成平衡二叉树。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

/**
 * 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) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        if(NULL==head) return NULL;
        int len = length(head);
        TreeNode *root=NULL;
        if(len==1) {
            root = createNode(head->val);
            return root;
        }
        int middle = len/2;
        ListNode *pre=middleNode(head,middle);
		ListNode *p=pre->next;
        root = createNode(p->val);
        ListNode *postHead=p->next;
		pre->next=NULL;
        p->next=NULL;
        root->left=sortedListToBST(head);
        root->right = sortedListToBST(postHead);
        return root;
    }
private:
    TreeNode *createNode(int val){
        return new TreeNode(val);
    }
    int length(ListNode *head){
        int len=0;
        while(head){
            ++len;
            head=head->next;
        }
        return len;
    }
    ListNode* middleNode(ListNode *head,int index){
        int i=0;
        ListNode *pre=NULL;
        while(i<index){
            pre=head;
            head=head->next;
            ++i;
        }
        return pre;
    }
};