Convert Sorted List to Binary Search Tree -- leetcodeGiven a singly linked list where elements are s

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

基本思路：

中序遍历的过程，与有序链表的顺序是一一对应的。

采用中序遍历构造进树的构造。

并利用取值范围，确定，根的位置，以及子树的范围。

故需要遍历整个链表，求得总的长度。

链表长度的一半，即为根结点所在位置。 左边则为左子树的取值范围，右边即为右子树的取值范围。

同上，递归应用到左子树，右子树。逐步缩范围，直到叶节点。

/**
 * 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) {
        int size = 0;
        ListNode *p = head;
        while (p) {
            ++size;
            p = p->next;
        }
        
        return helper(head, 0, size-1);
    }
    
    TreeNode *helper(ListNode *&head, int start, int stop) {
        if (start > stop) return 0;
        const int mid = start + (stop-start) / 2;
        TreeNode *left = helper(head, start, mid-1);
        TreeNode *root = new TreeNode(head->val);
        root->left = left;
        head = head->next;
        root->right = helper(head, mid+1, stop);
        return root;
    }
};

Convert Sorted List to Binary Search Tree -- leetcodeGiven a singly linked list where elements are s

原文：http://blog.csdn.net/elton_xiao/article/details/45599721