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