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:

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/**
* 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;
}
};