Convert Sorted List to Binary Search Tree

Desicription

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

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Given the sorted linked list: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5

Solution

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/**
* 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 {
private:
TreeNode* buildTree(int left, int right, vector<int>& nums) {
if(left > right)
return 0;
int mid = (left + right)>>1;
TreeNode* root = new TreeNode(nums[mid]);
root->left = buildTree(left, mid-1, nums);
root->right = buildTree(mid+1, right, nums);
return root;
}
public:
TreeNode* sortedListToBST(ListNode* head) {
vector<int> nums;
while(head) {
nums.push_back(head->val);
head = head->next;
}
return buildTree(0, nums.size()-1, nums);
}
};