Convert Sorted List to Balanced BST

Convert Sorted List to Balanced BST

Description:

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

Example:

      2
1->2->3 => /  
     1 3

分析：

最简单的办法是把链表变成一个vector，然后按照这个题目来就行了。

另一个方法是按照bottom-up的方式，把孩子节点挂到父亲节点上，这就需要递归实现，每次退出当前递归就把当前节点返回给父节点。代码如下。

参考：

Convert Sorted List to Balanced Binary Search Tree (BST)

Code：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: a tree node
     */
    TreeNode *sortedListToBST(ListNode *head) {
        // write your code here
        if (head == nullptr) return nullptr;
        ListNode *curr = head;
        
        int n = 0;
        while(curr != nullptr) {
            n++;
            curr = curr->next;
        }
        
        return sortedListToBST(head, 0, n-1);
    }
    
    TreeNode* sortedListToBST(ListNode *& list, int start, int end) {
        if (start > end) return NULL;
        
        // same as (start+end)/2, avoids overflow
        int mid = start + (end - start) / 2;
        
        TreeNode *leftChild = sortedListToBST(list, start, mid-1);
        TreeNode *parent = new TreeNode(list->val);
        parent->left = leftChild;
        
        list = list->next;
        parent->right = sortedListToBST(list, mid+1, end);
        
        return parent;
    }
};