LeetCode（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.

分析

给定有序链表，构造平衡的二叉查找树。

与上题本质相同，只不过采用了不同的数据结构，本题关键在于准确求取链表节点数，并计算根节点所在位置，正确划分左右子树的子链表。

注意：指针处理，避免出现空指针引用。

AC代码

/** * 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) {        if (head == NULL)            return NULL;        int size = 0;        ListNode *p = head;        while (p)        {            ++size;            p = p->next;        }        ListNode *l = head, *r = NULL;        //查找中间节点，用于构造二叉树根节点        ListNode *pre = head;        p = head;        int i = 0;        while (p && i < size / 2)        {            pre = p;            p = p->next;            ++i;        }        //p节点作为根节点        TreeNode *root = new TreeNode(p->val);        // 其余节点为右子树        r = p->next;        //之前节点为左子树        if (pre->next == p)            pre->next = NULL;        else            l = NULL;        root->left = sortedListToBST(l);        root->right = sortedListToBST(r);        return root;    }};

GitHub测试程序源码