【LeetCode】109. Convert Sorted List to Binary Search Tree

最新推荐文章于 2021-01-12 00:53:34 发布

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最新推荐文章于 2021-01-12 00:53:34 发布

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原文链接：http://www.cnblogs.com/ganganloveu/p/4061909.html

本文介绍了一种将已排序的链表转换为高度平衡二叉搜索树的方法，通过选取中间节点作为根节点，递归地构建左右子树，确保树的平衡。

Convert Sorted List to Binary Search Tree

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

为了满足平衡要求，容易想到提出中间节点作为树根，因为已排序，所以左右两侧天然满足BST的要求。

左右子串分别递归下去，上层根节点连接下层根节点即可完成。

/**
 * 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(head == NULL)
            return NULL;
        else if(head->next == NULL)
            return new TreeNode(head->val);
        ListNode* mid = findMid(head);
        TreeNode* root = new TreeNode(mid->val);
        root->left = sortedListToBST(head);
        root->right = sortedListToBST(mid->next);
        return root;
    }
    ListNode* findMid(ListNode* head)
    {
        ListNode* preslow = NULL;
        ListNode* slow = head;
        ListNode* fast = head;
        while(fast != NULL)
        {
            fast = fast->next;
            if(fast != NULL)
            {
                fast = fast->next;
                preslow = slow;
                slow = slow->next;
            }
        }
        //break the list into two parts
        preslow->next = NULL;
        return slow;
    }
};