[LeetCode]109. Convert Sorted List to Binary Search Tree

Description:

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

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Solution:

题意：将给定增序的单向链表转换成平衡二叉树。

思路：平衡二叉树的特性是左右子树高度差不超过1,。利用这个特性，我们可以使用递归的方法，每次“对半”搜索当前链表的正中间或靠左或靠右节点，将它作为这一层的根，然后将在它之前的的链表作为它的左子树，在它之后的链表作为它的右子树，不断重复递归即可。

/** * 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 length = 0;        for (ListNode* cur = head; cur != NULL; cur = cur->next)            length++;                return structChildTree(head, length);    }        TreeNode* structChildTree(ListNode* head, int length) {        // cout << head->val << "  " << length << endl;                ListNode* cur = head;        int curIndex = 0;        for (curIndex = 0; curIndex < length / 2; curIndex++) {            cur = cur->next;        }                TreeNode* root = new TreeNode(cur->val);        if (curIndex == 0)            root->left = NULL;        else            root->left = structChildTree(head, length / 2);                if (curIndex + 1 == length)            root->right = NULL;        else            root->right = structChildTree(cur->next, length - curIndex - 1);                return root;            }};