109. Convert Sorted List to Binary Search Tree

Given a singly linked list where elements are sorted in ascending order,

convert it to a height balanced BST.

==============

题目:将升序的链表,转化为BST(note:此BST要求是高度平衡,就是树种左右子树高度差不超过1)

思路:

def TreeNode* sortedListToBST(ListNode* head){

按照slow/fast方式找到链表的中间节点mid,

将升序链表在mid节点处切割成为两个链表head1,和head2

将中间节点的值,new一个新的TreeNode节点:TreeNode *root = new TreeNode(head2->val)

下面开始递归:
root的左子树就是sortedListToBST(head1)
root的右子树就是sortedListToBST(head2->next)
return root
}

代码如下:

/**
* 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==nullptr){
return nullptr;
}else if(head->next==nullptr){
return new TreeNode(head->val);
}

ListNode *slow,*fast,*prev;
slow = fast = head;
prev = nullptr;
while(fast && fast->next){
prev = slow;
slow = slow->next;
fast = fast->next->next;
}
prev->next = nullptr;
//showList(head);
//showList(slow);
TreeNode *root = new TreeNode(slow->val);
root->left = sortedListToBST(head);
root->right = sortedListToBST(slow->next);
return root;
}
};