109. Convert Sorted List to Binary Search Tree

Posted 2020-10-20 小河沟大河沟

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109. 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.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     /    -3   9
   /   /
 -10  5

解析

• 主要考察链表求中间节点
• 统一输入输出接口，需要将前后两段指针分开，也可以再加入一个尾指针参数
• 这个题是这几天做的最顺利的，一次AC掉

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        
        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* pre = NULL;
        if (!head)
        {
            return NULL;
        }
        if (!head->next)
        {
            TreeNode* temp = new TreeNode(head->val);
            return temp;
        }
        while (fast&&fast->next)
        {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        
        TreeNode* root = new TreeNode(slow->val);
        
        pre->next = NULL;
        slow = slow->next;
        root->left = sortedListToBST(head);
        root->right = sortedListToBST(slow);

        return root;
    }
};

题目来源

• 109. Convert Sorted List to Binary Search Tree